{"id":164,"date":"2019-04-05T20:59:33","date_gmt":"2019-04-06T00:59:33","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/bcitphys8400\/?post_type=chapter&#038;p=164"},"modified":"2019-04-12T18:51:46","modified_gmt":"2019-04-12T22:51:46","slug":"2-5-de-broglies-matter-waves","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/bcitphys8400\/chapter\/2-5-de-broglies-matter-waves\/","title":{"raw":"2.5 De Broglie\u2019s Matter Waves","rendered":"2.5 De Broglie\u2019s Matter Waves"},"content":{"raw":"<div data-type=\"abstract\" id=\"93831\" class=\"ui-has-child-title\"><header>\r\n<div class=\"textbox textbox--learning-objectives\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Learning Objectives<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Describe de Broglie\u2019s hypothesis of matter waves<\/li>\r\n \t<li>Explain how the de Broglie\u2019s hypothesis gives the rationale for the quantization of angular momentum in Bohr\u2019s quantum theory of the hydrogen atom<\/li>\r\n \t<li>Describe the Davisson\u2013Germer experiment<\/li>\r\n \t<li>Interpret de Broglie\u2019s idea of matter waves and how they account for electron diffraction phenomena<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<span style=\"font-size: 14pt\">Compton\u2019s formula established that an electromagnetic wave can behave like a particle of light when interacting with matter. In 1924, Louis<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span class=\"no-emphasis\" data-type=\"term\" id=\"term257\" style=\"font-size: 14pt\">de Broglie<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">proposed a new speculative hypothesis that electrons and other particles of matter can behave like waves. Today, this idea is known as<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span data-type=\"term\" id=\"term258\" style=\"font-size: 14pt\">de Broglie\u2019s hypothesis of matter waves<\/span><span style=\"font-size: 14pt\">. In 1926, De Broglie\u2019s hypothesis, together with Bohr\u2019s early quantum theory, led to the development of a new theory of<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span data-type=\"term\" id=\"term259\" style=\"font-size: 14pt\">wave quantum mechanics<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">to describe the physics of atoms and subatomic particles. Quantum mechanics has paved the way for new engineering inventions and technologies, such as the laser and magnetic resonance imaging (MRI). These new technologies drive discoveries in other sciences such as biology and chemistry.<\/span>\r\n\r\n<\/header><\/div>\r\n<p id=\"fs-id1163712336118\">According to de Broglie\u2019s hypothesis, massless photons as well as massive particles must satisfy one common set of relations that connect the energy<span>\u00a0<\/span><em data-effect=\"italics\">E<\/em><span>\u00a0<\/span>with the frequency<span>\u00a0<\/span><em data-effect=\"italics\">f<\/em>, and the linear momentum<span>\u00a0<\/span><em data-effect=\"italics\">p<\/em><span>\u00a0<\/span>with the wavelength<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-1-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb.<\/span><\/span><\/span><span>\u00a0<\/span>We have discussed these relations for photons in the context of Compton\u2019s effect. We are recalling them now in a more general context. Any particle that has energy and momentum is a<span>\u00a0<\/span><span data-type=\"term\" id=\"term260\">de Broglie wave<\/span><span>\u00a0<\/span>of frequency<span>\u00a0<\/span><em data-effect=\"italics\">f<\/em><span>\u00a0<\/span>and wavelength<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-2-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb:<\/span><\/span><\/span><\/p>\r\n\r\n<div class=\"equation-callout\" data-type=\"note\" id=\"fs-id1163712274996\"><section>\r\n<div data-type=\"equation\" id=\"fs-id1163712039504\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-3-Frame\">\r\n<div class=\"textbox\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-3-Frame\"><span class=\"MathJax_MathContainer\"><span>E=hf<\/span><\/span><\/div>\r\n<div class=\"os-equation-number\"><span class=\"os-number\">[2.53]<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<div class=\"equation-callout\" data-type=\"note\" id=\"fs-id1163712029396\"><section>\r\n<div data-type=\"equation\" id=\"fs-id1163711940183\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-4-Frame\">\r\n<div class=\"textbox\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-4-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb=hp.<\/span><\/span><\/div>\r\n<div class=\"os-equation-number\"><span class=\"os-number\">[2.54]<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<p id=\"fs-id1163712430653\">Here,<span>\u00a0<\/span><em data-effect=\"italics\">E<\/em><span>\u00a0<\/span>and<span>\u00a0<\/span><em data-effect=\"italics\">p<\/em><span>\u00a0<\/span>are, respectively, the relativistic energy and the momentum of a particle. De Broglie\u2019s relations are usually expressed in terms of the wave vector<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-5-Frame\"><span class=\"MathJax_MathContainer\"><span>k\u2192,<\/span><\/span><\/span><span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-6-Frame\"><span class=\"MathJax_MathContainer\"><span>k=2\u03c0\/\u03bb,<\/span><\/span><\/span><span>\u00a0<\/span>and the wave frequency<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-7-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03c9=2\u03c0f,<\/span><\/span><\/span><span>\u00a0<\/span>as we usually do for waves:<\/p>\r\n\r\n<div data-type=\"equation\" id=\"fs-id1163712146534\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-8-Frame\">\r\n<div class=\"textbox\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-8-Frame\"><span class=\"MathJax_MathContainer\"><span>E=\u210f\u03c9<\/span><\/span><\/div>\r\n<div class=\"os-equation-number\"><span class=\"os-number\">[2.55]<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"equation\" id=\"fs-id1163710848706\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-9-Frame\">\r\n<div class=\"textbox\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-9-Frame\"><span class=\"MathJax_MathContainer\"><span>p\u2192=\u210fk\u2192.<\/span><\/span><\/div>\r\n<div class=\"os-equation-number\"><span class=\"os-number\">[2.56]<\/span><\/div>\r\n<\/div>\r\n<span style=\"font-size: 14pt\">Wave theory tells us that a wave carries its energy with the<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span data-type=\"term\" id=\"term261\" style=\"font-size: 14pt\">group velocity<\/span><span style=\"font-size: 14pt\">. For matter waves, this group velocity is the velocity<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><em style=\"font-size: 14pt\" data-effect=\"italics\">u<\/em><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">of the particle. Identifying the energy<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><em style=\"font-size: 14pt\" data-effect=\"italics\">E<\/em><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">and momentum<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><em style=\"font-size: 14pt\" data-effect=\"italics\">p<\/em><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">of a particle with its relativistic energy<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-10-Frame\" style=\"font-size: 14pt\"><span class=\"MathJax_MathContainer\">mc2<\/span><\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">and its relativistic momentum<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><em style=\"font-size: 14pt\" data-effect=\"italics\">mu<\/em><span style=\"font-size: 14pt\">, respectively, it follows from de Broglie relations that matter waves satisfy the following relation:<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-11-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bbf=\u03c9k=E\/\u210fp\/\u210f=Ep=mc2mu=c2u=c\u03b2<\/span><\/span><\/div>\r\n<div class=\"os-equation-number\"><span class=\"os-number\">[2.57]<\/span><\/div>\r\n<\/div>\r\n<p id=\"fs-id1163711912925\">where<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-12-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03b2=u\/c.<\/span><\/span><\/span><span>\u00a0<\/span>When a particle is massless we have<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-13-Frame\"><span class=\"MathJax_MathContainer\"><span>u=c<\/span><\/span><\/span><span>\u00a0<\/span>and<span>\u00a0<\/span>Equation 2.57<span>\u00a0<\/span>becomes<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-14-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bbf=c.<\/span><\/span><\/span><\/p>\r\n\r\n<div data-type=\"example\" id=\"fs-id1163712438153\" class=\"ui-has-child-title\"><header><\/header><section>\r\n<div class=\"textbox shaded\"><header>\r\n<h3 class=\"os-title\"><span class=\"os-title-label\">EXAMPLE<span>\u00a02<\/span><\/span><span class=\"os-number\">.11<\/span><span class=\"os-divider\"><\/span><\/h3>\r\n<\/header><section>\r\n<p id=\"fs-id1172102060513\"><span data-type=\"title\"><strong>How Long Are de Broglie Matter Waves?<\/strong><\/span><\/p>\r\nCalculate the de Broglie wavelength of: (a) a 0.65-kg basketball thrown at a speed of 10 m\/s, (b) a nonrelativistic electron with a kinetic energy of 1.0 eV, and (c) a relativistic electron with a kinetic energy of<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-15-Frame\"><span class=\"MathJax_MathContainer\"><span>108ke V.<\/span><\/span><\/span>\r\n\r\n<span data-type=\"title\"><strong>Strategy\r\n<\/strong><\/span>We use<span>\u00a0<\/span>Equation 2.57<span>\u00a0<\/span>to find the de Broglie wavelength. When the problem involves a nonrelativistic object moving with a nonrelativistic speed<span>\u00a0<\/span><em data-effect=\"italics\">u<\/em>, such as in (a) when<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-16-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03b2=u\/c\u226a1,<\/span><\/span><\/span><span>\u00a0<\/span>we use nonrelativistic momentum<span>\u00a0<\/span><em data-effect=\"italics\">p<\/em>. When the nonrelativistic approximation cannot be used, such as in (c), we must use the relativistic momentum<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-17-Frame\"><span class=\"MathJax_MathContainer\"><span>p=mu=m0\u03b3u=E0\u03b3\u03b2,<\/span><\/span><\/span><span>\u00a0<\/span>where the rest mass energy of a particle is<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-18-Frame\"><span class=\"MathJax_MathContainer\"><span>E0=m0c2<\/span><\/span><\/span><span>\u00a0<\/span>and<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-19-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03b3<\/span><\/span><\/span><span>\u00a0<\/span>is the Lorentz factor<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-20-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03b3=1\/1\u2212\u03b22.<\/span><\/span><\/span><span>\u00a0<\/span>The total energy<span>\u00a0<\/span><em data-effect=\"italics\">E<\/em><span>\u00a0<\/span>of a particle is given by<span>\u00a0<\/span>Equation 2.53<span>\u00a0<\/span>and the kinetic energy is<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-21-Frame\"><span class=\"MathJax_MathContainer\"><span>K=E\u2212E0=(\u03b3\u22121)E0.<\/span><\/span><\/span><span>\u00a0<\/span>When the kinetic energy is known, we can invert<span>\u00a0<\/span>Equation 2.18<span>\u00a0<\/span>to find the momentum<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-22-Frame\"><span class=\"MathJax_MathContainer\"><span>p=(E2\u2212E02)\/c2=K(K+2E0)\/c<\/span><\/span><\/span><span>\u00a0<\/span>and substitute in<span>\u00a0<\/span>Equation 2.57<span>\u00a0<\/span>to obtain\r\n<div data-type=\"equation\" id=\"fs-id1163712341589\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-23-Frame\">\r\n<div class=\"textbox\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-23-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb=hp=hcK(K+2E0).<\/span><\/span><\/div>\r\n<div class=\"os-equation-number\">\r\n\r\n<span class=\"os-number\">[2.58]<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<span style=\"text-indent: 1em;font-size: 1rem\">Depending on the problem at hand, in this equation we can use the following values for<\/span><span style=\"text-indent: 1em;font-size: 1rem\">\u00a0<\/span><em style=\"text-indent: 1em;font-size: 1rem\" data-effect=\"italics\">hc<\/em><span style=\"text-indent: 1em;font-size: 1rem\">:<\/span><span style=\"text-indent: 1em;font-size: 1rem\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-24-Frame\" style=\"text-indent: 1em;font-size: 1rem\"><span class=\"MathJax_MathContainer\">hc=(6.626\u00d710\u221234J\u00b7s)(2.998\u00d7108m\/s)=1.986\u00d710\u221225J\u00b7m=1.241eV\u00b7\u03bcm<\/span><\/span>\r\n\r\n<\/div>\r\n<div class=\"os-equation-number\">\r\n\r\n<strong><span style=\"text-indent: 1em;font-size: 1rem\">Solution<\/span><\/strong>\r\n\r\n<\/div>\r\n<\/div>\r\n<ol id=\"fs-id1172101985706\" type=\"a\">\r\n \t<li>For the basketball, the kinetic energy is\r\n<span data-type=\"newline\">\r\n<\/span>\r\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163711937451\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-25-Frame\"><span class=\"MathJax_MathContainer\"><span>K=m0u2\/2=(0.65kg)(10m\/s)2\/2=32.5J<\/span><\/span><\/div>\r\n<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>and the rest mass energy is<span data-type=\"newline\">\r\n<\/span>\r\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163711938219\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-26-Frame\"><span class=\"MathJax_MathContainer\"><span>E0=m0c2=(0.65kg)(2.998\u00d7108m\/s)2=5.84\u00d71016J.<\/span><\/span><\/div>\r\n<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>We see that<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-27-Frame\"><span class=\"MathJax_MathContainer\"><span>K\/(K+E0)\u226a1and usep=m0u=(0.65kg)(10m\/s)=6.5J\u00b7s\/m:<\/span><\/span><\/span><span data-type=\"newline\">\r\n<\/span>\r\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163710827302\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-28-Frame\">\r\n\r\n<span class=\"MathJax_MathContainer\"><span>\u03bb=hp=6.626\u00d710\u221234J\u00b7s6.5J\u00b7s\/m=1.02\u00d710\u221234m.<\/span><\/span>\r\n\r\n<\/div>\r\n<\/div><\/li>\r\n \t<li>For the nonrelativistic electron,\r\n<span data-type=\"newline\">\r\n<\/span>\r\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163712317187\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-29-Frame\"><span class=\"MathJax_MathContainer\"><span>E0=m0c2=(9.109\u00d710\u221231kg)(2.998\u00d7108m\/s)2=511keV<\/span><\/span><\/div>\r\n<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>and when<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-30-Frame\"><span class=\"MathJax_MathContainer\"><span>K=1.0eV,<\/span><\/span><\/span><span>\u00a0<\/span>we have<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-31-Frame\"><span class=\"MathJax_MathContainer\"><span>K\/(K+E0)=(1\/512)\u00d710\u22123\u226a1,<\/span><\/span><\/span><span>\u00a0<\/span>so we can use the nonrelativistic formula. However, it is simpler here to use<span>\u00a0<\/span>Equation 2.58:<span data-type=\"newline\">\r\n<\/span>\r\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163712141043\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-32-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb=hp=hcK(K+2E0)=1.241eV\u00b7\u03bcm(1.0eV)[1.0eV+2(511keV)]=1.23nm.<\/span><\/span><\/div>\r\n<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>If we use nonrelativistic momentum, we obtain the same result because 1 eV is much smaller than the rest mass of the electron.<\/li>\r\n \t<li>For a fast electron with<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-33-Frame\"><span class=\"MathJax_MathContainer\"><span>K=108keV,<\/span><\/span><\/span><span>\u00a0<\/span>relativistic effects cannot be neglected because its total energy is<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-34-Frame\"><span class=\"MathJax_MathContainer\"><span>E=K+E0=108keV+511keV=619keV<\/span><\/span><\/span><span>\u00a0<\/span>and<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-35-Frame\"><span class=\"MathJax_MathContainer\"><span>K\/E=108\/619<\/span><\/span><\/span><span>\u00a0<\/span>is not negligible:\r\n<span data-type=\"newline\">\r\n<\/span>\r\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163712383831\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-36-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb=hp=hcK(K+2E0)=1.241eV\u00b7\u03bcm108keV[108keV+2(511keV)]=3.55pm.<\/span><\/span><\/div>\r\n<div><\/div>\r\n<div class=\"MathJax_MathML\">\r\n\r\n<span data-type=\"title\" style=\"text-indent: 1em;font-size: 1rem\"><strong>Significance<\/strong><\/span>\r\n\r\n<span style=\"text-indent: 1em;font-size: 1rem\">We see from these estimates that De Broglie\u2019s wavelengths of macroscopic objects such as a ball are immeasurably small. Therefore, even if they exist, they are not detectable and do not affect the motion of macroscopic objects.<\/span>\r\n\r\n<\/div>\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<div class=\"check-understanding ui-has-child-title\" data-type=\"note\" id=\"fs-id1163712281992\"><header><\/header><section>\r\n<div class=\"os-note-body\">\r\n<div class=\"os-hasSolution unnumbered\" data-type=\"exercise\" id=\"fs-id1163711954901\"><section>\r\n<div data-type=\"problem\" id=\"fs-id1163712212556\"><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<div class=\"textbox textbox--key-takeaways\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\"><span class=\"os-title-label\">CHECK YOUR UNDERSTANDING<span>\u00a02<\/span><\/span><span class=\"os-number\">.11<\/span><\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\"><header><span style=\"font-size: 1rem\">What is de Broglie\u2019s wavelength of a nonrelativistic proton with a kinetic energy of 1.0 eV?<\/span><\/header><\/div>\r\n<\/div>\r\n<p id=\"fs-id1163711963974\">Using the concept of the electron matter wave, de Broglie provided a rationale for the quantization of the electron\u2019s angular momentum in the hydrogen atom, which was postulated in Bohr\u2019s quantum theory. The physical explanation for the first Bohr quantization condition comes naturally when we assume that an electron in a hydrogen atom behaves not like a particle but like a wave. To see it clearly, imagine a stretched guitar string that is clamped at both ends and vibrates in one of its normal modes. If the length of the string is<span>\u00a0<\/span><em data-effect=\"italics\">l\u00a0<\/em>(Figure 2.18), the wavelengths of these vibrations cannot be arbitrary but must be such that an integer<span>\u00a0<\/span><em data-effect=\"italics\">k<\/em><span>\u00a0<\/span>number of half-wavelengths<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-37-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb\/2<\/span><\/span><\/span>fit exactly on the distance<span>\u00a0<\/span><em data-effect=\"italics\">l<\/em><span>\u00a0<\/span>between the ends. This is the condition<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-38-Frame\"><span class=\"MathJax_MathContainer\"><span>l=k\u03bb\/2<\/span><\/span><\/span><span>\u00a0<\/span>for a standing wave on a string. Now suppose that instead of having the string clamped at the walls, we bend its length into a circle and fasten its ends to each other. This produces a circular string that vibrates in normal modes, satisfying the same standing-wave condition, but the number of half-wavelengths must now be an even number<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-39-Frame\"><span class=\"MathJax_MathContainer\"><span>k,k=2n,<\/span><\/span><\/span><span>\u00a0<\/span>and the length<span>\u00a0<\/span><em data-effect=\"italics\">l<\/em><span>\u00a0<\/span>is now connected to the radius<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-40-Frame\"><span class=\"MathJax_MathContainer\"><span>rn<\/span><\/span><\/span><span>\u00a0<\/span>of the circle. This means that the radii are not arbitrary but must satisfy the following standing-wave condition:<\/p>\r\n\r\n<div data-type=\"equation\" id=\"fs-id1163711989563\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-41-Frame\">\r\n<div class=\"textbox\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-41-Frame\"><span class=\"MathJax_MathContainer\"><span>2\u03c0rn=2n\u03bb2.<\/span><\/span><\/div>\r\n<div class=\"os-equation-number\"><span class=\"os-number\">[2.59]<\/span><\/div>\r\n<\/div>\r\n<span style=\"font-size: 14pt\">If an electron in the<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><em style=\"font-size: 14pt\" data-effect=\"italics\">n<\/em><span style=\"font-size: 14pt\">th Bohr orbit moves as a wave, by<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">Equation 2.59<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">its wavelength must be equal to<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-42-Frame\" style=\"font-size: 14pt\"><span class=\"MathJax_MathContainer\">\u03bb=2\u03c0rn\/n.<\/span><\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">Assuming that<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">Equation 2.58<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">is valid, the electron wave of this wavelength corresponds to the electron\u2019s linear momentum,<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-43-Frame\" style=\"font-size: 14pt\"><span class=\"MathJax_MathContainer\">p=h\/\u03bb=nh\/(2\u03c0rn)=n\u210f\/rn.<\/span><\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">In a circular orbit, therefore, the electron\u2019s angular momentum must be<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"equation\" id=\"fs-id1163712425771\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-44-Frame\">\r\n<div class=\"textbox\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-44-Frame\"><span class=\"MathJax_MathContainer\"><span>Ln=rnp=rnn\u210frn=n\u210f.<\/span><\/span><\/div>\r\n<div class=\"os-equation-number\"><span class=\"os-number\">[2.60]<\/span><\/div>\r\n<\/div>\r\n<span style=\"font-size: 14pt\">This equation is the first of Bohr\u2019s quantization conditions, given by<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">Equation 2.36. Providing a physical explanation for Bohr\u2019s quantization condition is a convincing theoretical argument for the existence of matter waves.<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"os-figure\">\r\n<figure id=\"CNX_UPhysics_39_05_string\">\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"996\"]<img alt=\"Figure A is the standing-wave pattern for a string clamped in the wall. The distance between each node corresponds to the half gamma. Figure B is the standing-wave pattern for an electron wave trapped in the third Bohr orbit in the hydrogen atom. The wave has a circular shape with six nodes. The distance between each two node corresponds to the gamma.\" data-media-type=\"image\/jpeg\" id=\"27810\" src=\"https:\/\/cnx.org\/resources\/9959a8324ba37cdbf28a48d785a8d8787b4669b6\" width=\"996\" height=\"295\" \/> Figure 2.18 Standing-wave pattern: (a) a stretched string clamped at the walls; (b) an electron wave trapped in the third Bohr orbit in the hydrogen atom.[\/caption]<\/figure>\r\n<\/div>\r\n<div data-type=\"example\" id=\"fs-id1163710819156\" class=\"ui-has-child-title\"><header><\/header><section>\r\n<div class=\"textbox shaded\"><header>\r\n<h3 class=\"os-title\"><span class=\"os-title-label\">EXAMPLE<span>\u00a02<\/span><\/span><span class=\"os-number\">.12<\/span><span class=\"os-divider\"><\/span><\/h3>\r\n<\/header><section>\r\n<p id=\"fs-id1172099500868\"><span data-type=\"title\"><strong>The Electron Wave in the Ground State of Hydrogen<\/strong><\/span><\/p>\r\nFind the de Broglie wavelength of an electron in the ground state of hydrogen.\r\n\r\n<span data-type=\"title\"><strong>Strategy<\/strong><\/span>\r\n\r\nWe combine the first quantization condition in<span>\u00a0<\/span>Equation 2.60<span>\u00a0<\/span>with<span>\u00a0<\/span>Equation 2.36<span>\u00a0<\/span>and use<span>\u00a0<\/span>Equation 2.38<span>\u00a0<\/span>for the first Bohr radius with<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-45-Frame\"><span class=\"MathJax_MathContainer\"><span>n=1.<\/span><\/span><\/span>\r\n\r\n<span data-type=\"title\"><strong>Solution<\/strong><\/span>\r\n\r\nWhen<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-46-Frame\"><span class=\"MathJax_MathContainer\"><span>n=1<\/span><\/span><\/span><span>\u00a0<\/span>and<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-47-Frame\"><span class=\"MathJax_MathContainer\"><span>rn=a0=0.529\u00c5,<\/span><\/span><\/span><span>\u00a0<\/span>the Bohr quantization condition gives<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-48-Frame\"><span class=\"MathJax_MathContainer\"><span>a0p=1\u00b7\u210f\u21d2p=\u210f\/a0.<\/span><\/span><\/span><span>\u00a0<\/span>The electron wavelength is:\r\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163710983371\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-49-Frame\">\r\n\r\n<span class=\"MathJax_MathContainer\"><span>\u03bb=h\/p=h\/\u210f\/a0=2\u03c0a0=2\u03c0(0.529\u00c5)=3.324\u00c5.<\/span><\/span>\r\n\r\n<span data-type=\"title\" style=\"text-indent: 1em;font-size: 1rem\"><strong>Significance<\/strong><\/span>\r\n\r\n<span style=\"text-indent: 1em;font-size: 1rem\">We obtain the same result when we use<\/span><span style=\"text-indent: 1em;font-size: 1rem\">\u00a0<\/span>Equation 2.58<span style=\"text-indent: 1em;font-size: 1rem\">\u00a0<\/span><span style=\"text-indent: 1em;font-size: 1rem\">directly.<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<div class=\"textbox textbox--key-takeaways\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\"><span class=\"os-title-label\">CHECK YOUR UNDERSTANDING<span>\u00a02<\/span><\/span><span class=\"os-number\">.12<\/span><\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\"><header><span style=\"font-size: 1rem\">Find the de Broglie wavelength of an electron in the third excited state of hydrogen.<\/span><\/header><\/div>\r\n<\/div>\r\n<p id=\"fs-id1163712285476\">Experimental confirmation of matter waves came in 1927 when C.<span>\u00a0<\/span><span class=\"no-emphasis\" data-type=\"term\" id=\"term262\">Davisson<\/span><span>\u00a0<\/span>and L.<span>\u00a0<\/span><span class=\"no-emphasis\" data-type=\"term\" id=\"term263\">Germer<\/span><span>\u00a0<\/span>performed a series of electron-scattering experiments that clearly showed that electrons do behave like waves. Davisson and Germer did not set up their experiment to confirm de Broglie\u2019s hypothesis: The confirmation came as a byproduct of their routine experimental studies of metal surfaces under electron bombardment.<\/p>\r\n<p id=\"fs-id1163712032514\">In the particular experiment that provided the very first evidence of electron waves (known today as the<span>\u00a0<\/span><span data-type=\"term\" id=\"term264\">Davisson\u2013Germer experiment<\/span>), they studied a surface of nickel. Their nickel sample was specially prepared in a high-temperature oven to change its usual polycrystalline structure to a form in which large single-crystal domains occupy the volume.<span>\u00a0<\/span>Figure 2.19<span>\u00a0<\/span>shows the experimental setup. Thermal electrons are released from a heated element (usually made of tungsten) in the electron gun and accelerated through a potential difference<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-50-Frame\"><span class=\"MathJax_MathContainer\"><span>\u0394V,<\/span><\/span><\/span><span>\u00a0<\/span>becoming a well-collimated beam of electrons produced by an electron gun. The kinetic energy<span>\u00a0<\/span><em data-effect=\"italics\">K<\/em><span>\u00a0<\/span>of the electrons is adjusted by selecting a value of the potential difference in the electron gun. This produces a beam of electrons with a set value of linear momentum, in accordance with the conservation of energy:<\/p>\r\n\r\n<div data-type=\"equation\" id=\"fs-id1163711981923\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-51-Frame\">\r\n<div class=\"textbox\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-51-Frame\"><span class=\"MathJax_MathContainer\"><span>e\u0394V=K=p22m\u21d2p=2me\u0394V.<\/span><\/span><\/div>\r\n<div class=\"os-equation-number\"><span class=\"os-number\">[2.61]<\/span><\/div>\r\n<\/div>\r\n<span style=\"font-size: 14pt\">The electron beam is incident on the nickel sample in the direction normal to its surface. At the surface, it scatters in various directions. The intensity of the beam scattered in a selected direction<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-52-Frame\" style=\"font-size: 14pt\"><span class=\"MathJax_MathContainer\">\u03c6<\/span><\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">is measured by a highly sensitive detector. The detector\u2019s angular position with respect to the direction of the incident beam can be varied from<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-53-Frame\" style=\"font-size: 14pt\"><span class=\"MathJax_MathContainer\">\u03c6=0\u00b0<\/span><\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">to<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-54-Frame\" style=\"font-size: 14pt\"><span class=\"MathJax_MathContainer\">\u03c6=90\u00b0.<\/span><\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">The entire setup is enclosed in a vacuum chamber to prevent electron collisions with air molecules, as such thermal collisions would change the electrons\u2019 kinetic energy and are not desirable.<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"os-figure\">\r\n<figure id=\"CNX_UPhysics_39_05_davisson\">\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"457\"]<img alt=\"Figure shows the schematics of the experimental setup of the Davisson\u2013Germer diffraction experiment. A beam of electrons is emitted by the electron gun, passes through the collimator, and hits Nickel target. Diffracted beam forms an angle phi with the incident beam and is detected by a moving detector. All of this is shown happening in a vacuum\" data-media-type=\"image\/jpeg\" id=\"48111\" src=\"https:\/\/cnx.org\/resources\/8e4ffc6549ef9a8e810bde4b4ae984c3085a89d9\" width=\"457\" height=\"336\" \/> Figure 2.19 Schematics of the experimental setup of the Davisson\u2013Germer diffraction experiment. A well-collimated beam of electrons is scattered off the nickel target. The kinetic energy of electrons in the incident beam is selected by adjusting a variable potential, \u0394V, in the electron gun. Intensity of the scattered electron beam is measured for a range of scattering angles \u03c6, whereas the distance between the detector and the target does not change.[\/caption]<\/figure>\r\n<\/div>\r\n<p id=\"fs-id1163711938476\">When the nickel target has a polycrystalline form with many randomly oriented microscopic crystals, the incident electrons scatter off its surface in various random directions. As a result, the intensity of the scattered electron beam is much the same in any direction, resembling a diffuse reflection of light from a porous surface. However, when the nickel target has a regular crystalline structure, the intensity of the scattered electron beam shows a clear maximum at a specific angle and the results show a clear diffraction pattern (see<span>\u00a0<\/span>Figure 2.20). Similar diffraction patterns formed by X-rays scattered by various crystalline solids were studied in 1912 by father-and-son physicists William H.<span>\u00a0<\/span><span class=\"no-emphasis\" data-type=\"term\" id=\"term265\">Bragg<\/span><span>\u00a0<\/span>and William L.<span>\u00a0<\/span><span class=\"no-emphasis\" data-type=\"term\" id=\"term266\">Bragg<\/span>. The Bragg law in X-ray crystallography provides a connection between the wavelength<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-57-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb<\/span><\/span><\/span><span>\u00a0<\/span>of the radiation incident on a crystalline lattice, the lattice spacing, and the position of the interference maximum in the diffracted radiation (see<span>\u00a0<\/span><a class=\"target-chapter\" href=\"https:\/\/cnx.org\/contents\/af275420-6050-4707-995c-57b9cc13c358@11.1:530d811a-3df9-4913-9c8b-5844eae74384@5\" data-page=\"28\">Diffraction<\/a>).<\/p>\r\n<p id=\"fs-id1163710853034\">The lattice spacing of the Davisson\u2013Germer target, determined with X-ray crystallography, was measured to be<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-58-Frame\"><span class=\"MathJax_MathContainer\"><span>a=2.15\u00c5.<\/span><\/span><\/span><span>\u00a0<\/span>Unlike X-ray crystallography in which X-rays penetrate the sample, in the original Davisson\u2013Germer experiment, only the surface atoms interact with the incident electron beam. For the surface diffraction, the maximum intensity of the reflected electron beam is observed for scattering angles that satisfy the condition<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-59-Frame\"><span class=\"MathJax_MathContainer\"><span>n\u03bb=asin\u03c6<\/span><\/span><\/span><span>\u00a0<\/span>(see<span>\u00a0<\/span>Figure 2.21). The first-order maximum (for<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-60-Frame\"><span class=\"MathJax_MathContainer\"><span>n=1<\/span><\/span><\/span>) is measured at a scattering angle of<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-61-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03c6\u224850\u00b0<\/span><\/span><\/span><span>\u00a0<\/span>at<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-62-Frame\"><span class=\"MathJax_MathContainer\"><span>\u0394V\u224854V,<\/span><\/span><\/span><span>\u00a0<\/span>which gives the wavelength of the incident radiation as<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-63-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb=(2.15\u00c5)sin50\u00b0=1.64\u00c5.<\/span><\/span><\/span><span>\u00a0<\/span>On the other hand, a 54-V potential accelerates the incident electrons to kinetic energies of<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-64-Frame\"><span class=\"MathJax_MathContainer\"><span>K=54eV.<\/span><\/span><\/span><span>\u00a0<\/span>Their momentum, calculated from<span>\u00a0<\/span>Equation 2.61, is<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-65-Frame\"><span class=\"MathJax_MathContainer\"><span>p=2.478\u00d710\u22125eV\u00b7s\/m.<\/span><\/span><\/span><span>\u00a0<\/span>When we substitute this result in<span>\u00a0<\/span>Equation 2.58, the de Broglie wavelength is obtained as<\/p>\r\n\r\n<div data-type=\"equation\" id=\"fs-id1163712085202\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-66-Frame\">\r\n<div class=\"textbox\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-66-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb=hp=4.136\u00d710\u221215eV\u00b7s2.478\u00d710\u22125eV\u00b7s\/m=1.67\u00c5.<\/span><\/span><\/div>\r\n<div class=\"os-equation-number\"><span class=\"os-number\">[2.62]<\/span><\/div>\r\n<\/div>\r\n<span style=\"font-size: 14pt\">The same result is obtained when we use<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-67-Frame\" style=\"font-size: 14pt\"><span class=\"MathJax_MathContainer\">K=54eV<\/span><\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">in<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">Equation 2.61. The proximity of this theoretical result to the Davisson\u2013Germer experimental value of<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-68-Frame\" style=\"font-size: 14pt\"><span class=\"MathJax_MathContainer\">\u03bb=1.64\u00c5<\/span><\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">is a convincing argument for the existence of de Broglie matter waves.<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"os-figure\">\r\n<figure id=\"CNX_UPhysics_39_05_davisson2\">\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"347\"]<img alt=\"The graph shows the dependence of the intensity of the scattering beam on the scattering angle in degrees. The intensity degrees from 10 to 30 degrees, followed by a sharp increase and maximum at 50 degrees, and then reaches zero at 80 degrees.\" data-media-type=\"image\/jpeg\" id=\"21884\" src=\"https:\/\/cnx.org\/resources\/822ba59021544c10997c333c09a9007ff22072b5\" width=\"347\" height=\"346\" \/> Figure 2.20 The experimental results of electron diffraction on a nickel target for the accelerating potential in the electron gun of about \u0394V=54V: The intensity maximum is registered at the scattering angle of about \u03c6=50\u00b0.[\/caption]\r\n\r\n&nbsp;<\/figure>\r\n<\/div>\r\n<div class=\"os-figure\">\r\n<figure id=\"CNX_UPhysics_39_05_bragg\">\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"333\"]<img alt=\"Figure shows the surface diffraction of a monochromatic electromagnetic wave on a crystalline lattice structure. The in-phase incident beams are reflected from atoms on the surface. Phi is the angle between the incident and the reflected beam, the in-plane distance between the atoms is a.\" data-media-type=\"image\/jpeg\" id=\"28527\" src=\"https:\/\/cnx.org\/resources\/b9cd5291754f7d099d4511e8045535dd97197ce6\" width=\"333\" height=\"249\" \/> Figure 2.21 In the surface diffraction of a monochromatic electromagnetic wave on a crystalline lattice structure, the in-phase incident beams are reflected from atoms on the surface. A ray reflected from the left atom travels an additional distance D=asin\u03c6 to the detector, where a is the lattice spacing. The reflected beams remain in-phase when D is an integer multiple of their wavelength \u03bb. The intensity of the reflected waves has pronounced maxima for angles \u03c6 satisfying n\u03bb=asin\u03c6.[\/caption]<\/figure>\r\n<\/div>\r\n<p id=\"fs-id1163711996022\">Diffraction lines measured with low-energy electrons, such as those used in the Davisson\u2013Germer experiment, are quite broad (see<span>\u00a0<\/span>Figure 2.20) because the incident electrons are scattered only from the surface. The resolution of diffraction images greatly improves when a higher-energy electron beam passes through a thin metal foil. This occurs because the diffraction image is created by scattering off many crystalline planes inside the volume, and the maxima produced in scattering at Bragg angles are sharp (see<span>\u00a0<\/span>Figure 2.22).<\/p>\r\n\r\n<div class=\"os-figure\">\r\n<figure id=\"CNX_UPhysics_39_05_laue\">\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"975\"]<img alt=\"Picture A is a photograph of the diffraction pattern obtained in scattering on a crystalline solid with X-rays. Picture B is a photograph of the diffraction pattern obtained in scattering on a crystalline solid with electrons. Both pictures demonstrate diffracted spots symmetrically arranged around the central beam.\" data-media-type=\"image\/jpeg\" id=\"13095\" src=\"https:\/\/cnx.org\/resources\/b261ff9f58dc5c1f3adbd87b482fbeb0839e5853\" width=\"975\" height=\"522\" \/> Figure 2.22 Diffraction patterns obtained in scattering on a crystalline solid: (a) with X-rays, and (b) with electrons. The observed pattern reflects the symmetry of the crystalline structure of the sample.<span class=\"os-caption\"><\/span>[\/caption]<\/figure>\r\n<\/div>\r\n<p id=\"fs-id1163711912089\">Since the work of Davisson and Germer, de Broglie\u2019s hypothesis has been extensively tested with various experimental techniques, and the existence of de Broglie waves has been confirmed for numerous elementary particles. Neutrons have been used in scattering experiments to determine crystalline structures of solids from interference patterns formed by neutron matter waves. The neutron has zero charge and its mass is comparable with the mass of a positively charged proton. Both neutrons and protons can be seen as matter waves. Therefore, the property of being a matter wave is not specific to electrically charged particles but is true of all particles in motion. Matter waves of molecules as large as carbon<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-75-Frame\"><span class=\"MathJax_MathContainer\"><span>C60<\/span><\/span><\/span><span>\u00a0<\/span>have been measured. All physical objects, small or large, have an associated matter wave as long as they remain in motion. The universal character of de Broglie matter waves is firmly established.<\/p>\r\n\r\n<div data-type=\"example\" id=\"fs-id1163712192182\" class=\"ui-has-child-title\"><header><\/header><section>\r\n<div class=\"textbox shaded\"><header>\r\n<h3 class=\"os-title\"><span class=\"os-title-label\">EXAMPLE<span>\u00a02<\/span><\/span><span class=\"os-number\">.13<\/span><span class=\"os-divider\"><\/span><\/h3>\r\n<\/header><section>\r\n<p id=\"fs-id1172102099642\"><span data-type=\"title\"><strong>Neutron Scattering<\/strong><\/span><\/p>\r\nSuppose that a neutron beam is used in a diffraction experiment on a typical crystalline solid. Estimate the kinetic energy of a neutron (in eV) in the neutron beam and compare it with kinetic energy of an ideal gas in equilibrium at room temperature.\r\n\r\n<span data-type=\"title\"><strong>Strategy<\/strong><\/span>\r\n\r\nWe assume that a typical crystal spacing<span>\u00a0<\/span><em data-effect=\"italics\">a<\/em><span>\u00a0<\/span>is of the order of 1.0 \u00c5. To observe a diffraction pattern on such a lattice, the neutron wavelength<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-76-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb<\/span><\/span><\/span><span>\u00a0<\/span>must be on the same order of magnitude as the lattice spacing. We use<span>\u00a0<\/span>Equation 2.61<span>\u00a0<\/span>to find the momentum<span>\u00a0<\/span><em data-effect=\"italics\">p<\/em><span>\u00a0<\/span>and kinetic energy<span>\u00a0<\/span><em data-effect=\"italics\">K<\/em>. To compare this energy with the energy<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-77-Frame\"><span class=\"MathJax_MathContainer\"><span>ET<\/span><\/span><\/span><span>\u00a0<\/span>of ideal gas in equilibrium at room temperature<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-78-Frame\"><span class=\"MathJax_MathContainer\"><span>T=300K,<\/span><\/span><\/span><span>\u00a0<\/span>we use the relation<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-79-Frame\"><span class=\"MathJax_MathContainer\"><span>K=32kBT,<\/span><\/span><\/span><span>\u00a0<\/span>where<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-80-Frame\"><span class=\"MathJax_MathContainer\"><span>kB=8.62\u00d710\u22125eV\/K<\/span><\/span><\/span><span>\u00a0<\/span>is the Boltzmann constant.\r\n\r\n<span data-type=\"title\"><strong>Solution<\/strong><\/span>\r\n\r\nWe evaluate<span>\u00a0<\/span><em data-effect=\"italics\">pc<\/em><span>\u00a0<\/span>to compare it with the neutron\u2019s rest mass energy<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-81-Frame\"><span class=\"MathJax_MathContainer\"><span>E0=940MeV:<\/span><\/span><\/span>\r\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163711912453\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-82-Frame\">\r\n\r\n<span class=\"MathJax_MathContainer\"><span>p=h\u03bb\u21d2pc=hc\u03bb=1.241\u00d710\u22126eV\u00b7m10\u221210m=12.41keV.<\/span><\/span>\r\n\r\n<span style=\"text-indent: 1em;font-size: 1rem\">We see that<\/span><span style=\"text-indent: 1em;font-size: 1rem\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-83-Frame\" style=\"text-indent: 1em;font-size: 1rem\"><span class=\"MathJax_MathContainer\">p2c2\u226aE02<\/span><\/span><span style=\"text-indent: 1em;font-size: 1rem\">\u00a0<\/span><span style=\"text-indent: 1em;font-size: 1rem\">so<\/span><span style=\"text-indent: 1em;font-size: 1rem\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-84-Frame\" style=\"text-indent: 1em;font-size: 1rem\"><span class=\"MathJax_MathContainer\">K\u226aE0<\/span><\/span><span style=\"text-indent: 1em;font-size: 1rem\">\u00a0<\/span><span style=\"text-indent: 1em;font-size: 1rem\">and we can use the nonrelativistic kinetic energy:<\/span>\r\n\r\n<span style=\"font-size: 1rem;text-indent: 0px\">K=p22mn=h22\u03bb2mn=(6.63\u00d710\u221234J\u00b7s)2(2\u00d710\u221220m2)(1.66\u00d710\u221227kg)=1.32\u00d710\u221220J=82.7meV.<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163711892811\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-85-Frame\">\r\n\r\n<span style=\"text-indent: 1em;font-size: 1rem\">Kinetic energy of ideal gas in equilibrium at 300 K is:<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163712283018\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-86-Frame\">\r\n\r\n<span class=\"MathJax_MathContainer\"><span class=\"MathJax_MathContainer\"><span>KT=32kBT=32(8.62\u00d710\u22125eV\/K)(300K)=38.8MeV.<\/span><\/span><\/span><span style=\"text-indent: 1em;font-size: 1rem\"><span style=\"text-indent: 1em;font-size: 1rem\"><span style=\"text-indent: 1em;font-size: 1rem\">We see that these energies are of the same order of magnitude.<\/span><\/span><\/span>\r\n\r\n<span data-type=\"title\" style=\"text-indent: 1em;font-size: 1rem\"><span data-type=\"title\" style=\"text-indent: 1em;font-size: 1rem\"><strong>Significance<\/strong><\/span><\/span>\r\n\r\n<span style=\"font-size: 1rem;text-indent: 1em\">Neutrons with energies in this range, which is typical for an ideal gas at room temperature, are called \u201cthermal neutrons.\u201d<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<div class=\"textbox shaded\"><header>\r\n<h3 class=\"os-title\"><span class=\"os-title-label\">EXAMPLE<span>\u00a02<\/span><\/span><span class=\"os-number\">.14<\/span><span class=\"os-divider\"><\/span><\/h3>\r\n<\/header><section>\r\n<p id=\"fs-id1172099743389\"><span data-type=\"title\"><strong>Wavelength of a Relativistic Proton<\/strong><\/span><\/p>\r\nIn a supercollider at CERN, protons can be accelerated to velocities of 0.75<em data-effect=\"italics\">c<\/em>. What are their de Broglie wavelengths at this speed? What are their kinetic energies?\r\n\r\n<span data-type=\"title\"><strong>Strategy<\/strong><\/span>\r\n\r\nThe rest mass energy of a proton is<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-87-Frame\"><span class=\"MathJax_MathContainer\"><span>E0=m0c2=(1.672\u00d710\u221227kg)(2.998\u00d7108m\/s)2=938MeV.<\/span><\/span><\/span><span>\u00a0<\/span>When the proton\u2019s velocity is known, we have<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-88-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03b2=0.75<\/span><\/span><\/span><span>\u00a0<\/span>and<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-89-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03b2\u03b3=0.75\/1\u22120.752=1.714.<\/span><\/span><\/span><span>\u00a0<\/span>We obtain the wavelength<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-90-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb<\/span><\/span><\/span><span>\u00a0<\/span>and kinetic energy<span>\u00a0<\/span><em data-effect=\"italics\">K<\/em><span>\u00a0<\/span>from relativistic relations.\r\n\r\n<strong>Solution<\/strong>\r\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163711940582\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-91-Frame\">\r\n\r\n<span class=\"MathJax_MathContainer\"><span>\u03bb=hp=hcpc=hc\u03b2\u03b3E0=1.241eV\u00b7\u03bcm1.714(938MeV)=0.77fm<\/span><\/span>\r\n\r\n<span style=\"font-size: 1rem;text-indent: 0px\">K=E0(\u03b3\u22121)=938MeV(1\/1\u22120.752\u22121)=480.1MeV<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163712167572\">\r\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-92-Frame\">\r\n\r\n<span data-type=\"title\" style=\"text-indent: 1em;font-size: 1rem\"><strong>Significance<\/strong><\/span>\r\n\r\n<span style=\"text-indent: 1em;font-size: 1rem\">Notice that because a proton is 1835 times more massive than an electron, if this experiment were performed with electrons, a simple rescaling of these results would give us the electron\u2019s wavelength of<\/span><span style=\"text-indent: 1em;font-size: 1rem\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-93-Frame\" style=\"text-indent: 1em;font-size: 1rem\"><span class=\"MathJax_MathContainer\">(1835)0.77fm=1.4pm<\/span><\/span><span style=\"text-indent: 1em;font-size: 1rem\">\u00a0<\/span><span style=\"text-indent: 1em;font-size: 1rem\">and its kinetic energy of<\/span><span style=\"text-indent: 1em;font-size: 1rem\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-94-Frame\" style=\"text-indent: 1em;font-size: 1rem\"><span class=\"MathJax_MathContainer\">480.1MeV\/1835=261.6keV.<\/span><\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<div class=\"textbox textbox--key-takeaways\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\"><span class=\"os-title-label\">CHECK YOUR UNDERSTANDING<span>\u00a02<\/span><\/span><span class=\"os-number\">.13<\/span><\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\"><header>\r\n<div class=\"os-title\"><span style=\"font-size: 1rem\">Find the de Broglie wavelength and kinetic energy of a free electron that travels at a speed of 0.75<\/span><em style=\"font-size: 1rem\" data-effect=\"italics\">c<\/em><span style=\"font-size: 1rem\">.<\/span><\/div>\r\n<\/header><\/div>\r\n<\/div>\r\n&nbsp;\r\n\r\n&nbsp;\r\n<div class=\"textbox\"><em>Download for free at http:\/\/cnx.org\/contents\/af275420-6050-4707-995c-57b9cc13c358@11.1<\/em><\/div>","rendered":"<div data-type=\"abstract\" id=\"93831\" class=\"ui-has-child-title\">\n<header>\n<div class=\"textbox textbox--learning-objectives\"><\/div>\n<\/header>\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Learning Objectives<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Describe de Broglie\u2019s hypothesis of matter waves<\/li>\n<li>Explain how the de Broglie\u2019s hypothesis gives the rationale for the quantization of angular momentum in Bohr\u2019s quantum theory of the hydrogen atom<\/li>\n<li>Describe the Davisson\u2013Germer experiment<\/li>\n<li>Interpret de Broglie\u2019s idea of matter waves and how they account for electron diffraction phenomena<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<p><span style=\"font-size: 14pt\">Compton\u2019s formula established that an electromagnetic wave can behave like a particle of light when interacting with matter. In 1924, Louis<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span class=\"no-emphasis\" data-type=\"term\" id=\"term257\" style=\"font-size: 14pt\">de Broglie<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">proposed a new speculative hypothesis that electrons and other particles of matter can behave like waves. Today, this idea is known as<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span data-type=\"term\" id=\"term258\" style=\"font-size: 14pt\">de Broglie\u2019s hypothesis of matter waves<\/span><span style=\"font-size: 14pt\">. In 1926, De Broglie\u2019s hypothesis, together with Bohr\u2019s early quantum theory, led to the development of a new theory of<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span data-type=\"term\" id=\"term259\" style=\"font-size: 14pt\">wave quantum mechanics<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">to describe the physics of atoms and subatomic particles. Quantum mechanics has paved the way for new engineering inventions and technologies, such as the laser and magnetic resonance imaging (MRI). These new technologies drive discoveries in other sciences such as biology and chemistry.<\/span><\/p>\n<p id=\"fs-id1163712336118\">According to de Broglie\u2019s hypothesis, massless photons as well as massive particles must satisfy one common set of relations that connect the energy<span>\u00a0<\/span><em data-effect=\"italics\">E<\/em><span>\u00a0<\/span>with the frequency<span>\u00a0<\/span><em data-effect=\"italics\">f<\/em>, and the linear momentum<span>\u00a0<\/span><em data-effect=\"italics\">p<\/em><span>\u00a0<\/span>with the wavelength<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-1-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb.<\/span><\/span><\/span><span>\u00a0<\/span>We have discussed these relations for photons in the context of Compton\u2019s effect. We are recalling them now in a more general context. Any particle that has energy and momentum is a<span>\u00a0<\/span><span data-type=\"term\" id=\"term260\">de Broglie wave<\/span><span>\u00a0<\/span>of frequency<span>\u00a0<\/span><em data-effect=\"italics\">f<\/em><span>\u00a0<\/span>and wavelength<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-2-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb:<\/span><\/span><\/span><\/p>\n<div class=\"equation-callout\" data-type=\"note\" id=\"fs-id1163712274996\">\n<section>\n<div data-type=\"equation\" id=\"fs-id1163712039504\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-3-Frame\">\n<div class=\"textbox\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-3-Frame\"><span class=\"MathJax_MathContainer\"><span>E=hf<\/span><\/span><\/div>\n<div class=\"os-equation-number\"><span class=\"os-number\">[2.53]<\/span><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<div class=\"equation-callout\" data-type=\"note\" id=\"fs-id1163712029396\">\n<section>\n<div data-type=\"equation\" id=\"fs-id1163711940183\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-4-Frame\">\n<div class=\"textbox\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-4-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb=hp.<\/span><\/span><\/div>\n<div class=\"os-equation-number\"><span class=\"os-number\">[2.54]<\/span><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<p id=\"fs-id1163712430653\">Here,<span>\u00a0<\/span><em data-effect=\"italics\">E<\/em><span>\u00a0<\/span>and<span>\u00a0<\/span><em data-effect=\"italics\">p<\/em><span>\u00a0<\/span>are, respectively, the relativistic energy and the momentum of a particle. De Broglie\u2019s relations are usually expressed in terms of the wave vector<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-5-Frame\"><span class=\"MathJax_MathContainer\"><span>k\u2192,<\/span><\/span><\/span><span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-6-Frame\"><span class=\"MathJax_MathContainer\"><span>k=2\u03c0\/\u03bb,<\/span><\/span><\/span><span>\u00a0<\/span>and the wave frequency<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-7-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03c9=2\u03c0f,<\/span><\/span><\/span><span>\u00a0<\/span>as we usually do for waves:<\/p>\n<div data-type=\"equation\" id=\"fs-id1163712146534\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-8-Frame\">\n<div class=\"textbox\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-8-Frame\"><span class=\"MathJax_MathContainer\"><span>E=\u210f\u03c9<\/span><\/span><\/div>\n<div class=\"os-equation-number\"><span class=\"os-number\">[2.55]<\/span><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"equation\" id=\"fs-id1163710848706\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-9-Frame\">\n<div class=\"textbox\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-9-Frame\"><span class=\"MathJax_MathContainer\"><span>p\u2192=\u210fk\u2192.<\/span><\/span><\/div>\n<div class=\"os-equation-number\"><span class=\"os-number\">[2.56]<\/span><\/div>\n<\/div>\n<p><span style=\"font-size: 14pt\">Wave theory tells us that a wave carries its energy with the<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span data-type=\"term\" id=\"term261\" style=\"font-size: 14pt\">group velocity<\/span><span style=\"font-size: 14pt\">. For matter waves, this group velocity is the velocity<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><em style=\"font-size: 14pt\" data-effect=\"italics\">u<\/em><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">of the particle. Identifying the energy<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><em style=\"font-size: 14pt\" data-effect=\"italics\">E<\/em><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">and momentum<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><em style=\"font-size: 14pt\" data-effect=\"italics\">p<\/em><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">of a particle with its relativistic energy<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-10-Frame\" style=\"font-size: 14pt\"><span class=\"MathJax_MathContainer\">mc2<\/span><\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">and its relativistic momentum<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><em style=\"font-size: 14pt\" data-effect=\"italics\">mu<\/em><span style=\"font-size: 14pt\">, respectively, it follows from de Broglie relations that matter waves satisfy the following relation:<\/span><\/p>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-11-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bbf=\u03c9k=E\/\u210fp\/\u210f=Ep=mc2mu=c2u=c\u03b2<\/span><\/span><\/div>\n<div class=\"os-equation-number\"><span class=\"os-number\">[2.57]<\/span><\/div>\n<\/div>\n<p id=\"fs-id1163711912925\">where<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-12-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03b2=u\/c.<\/span><\/span><\/span><span>\u00a0<\/span>When a particle is massless we have<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-13-Frame\"><span class=\"MathJax_MathContainer\"><span>u=c<\/span><\/span><\/span><span>\u00a0<\/span>and<span>\u00a0<\/span>Equation 2.57<span>\u00a0<\/span>becomes<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-14-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bbf=c.<\/span><\/span><\/span><\/p>\n<div data-type=\"example\" id=\"fs-id1163712438153\" class=\"ui-has-child-title\">\n<header><\/header>\n<section>\n<div class=\"textbox shaded\">\n<header>\n<h3 class=\"os-title\"><span class=\"os-title-label\">EXAMPLE<span>\u00a02<\/span><\/span><span class=\"os-number\">.11<\/span><span class=\"os-divider\"><\/span><\/h3>\n<\/header>\n<section>\n<p id=\"fs-id1172102060513\"><span data-type=\"title\"><strong>How Long Are de Broglie Matter Waves?<\/strong><\/span><\/p>\n<p>Calculate the de Broglie wavelength of: (a) a 0.65-kg basketball thrown at a speed of 10 m\/s, (b) a nonrelativistic electron with a kinetic energy of 1.0 eV, and (c) a relativistic electron with a kinetic energy of<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-15-Frame\"><span class=\"MathJax_MathContainer\"><span>108ke V.<\/span><\/span><\/span><\/p>\n<p><span data-type=\"title\"><strong>Strategy<br \/>\n<\/strong><\/span>We use<span>\u00a0<\/span>Equation 2.57<span>\u00a0<\/span>to find the de Broglie wavelength. When the problem involves a nonrelativistic object moving with a nonrelativistic speed<span>\u00a0<\/span><em data-effect=\"italics\">u<\/em>, such as in (a) when<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-16-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03b2=u\/c\u226a1,<\/span><\/span><\/span><span>\u00a0<\/span>we use nonrelativistic momentum<span>\u00a0<\/span><em data-effect=\"italics\">p<\/em>. When the nonrelativistic approximation cannot be used, such as in (c), we must use the relativistic momentum<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-17-Frame\"><span class=\"MathJax_MathContainer\"><span>p=mu=m0\u03b3u=E0\u03b3\u03b2,<\/span><\/span><\/span><span>\u00a0<\/span>where the rest mass energy of a particle is<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-18-Frame\"><span class=\"MathJax_MathContainer\"><span>E0=m0c2<\/span><\/span><\/span><span>\u00a0<\/span>and<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-19-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03b3<\/span><\/span><\/span><span>\u00a0<\/span>is the Lorentz factor<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-20-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03b3=1\/1\u2212\u03b22.<\/span><\/span><\/span><span>\u00a0<\/span>The total energy<span>\u00a0<\/span><em data-effect=\"italics\">E<\/em><span>\u00a0<\/span>of a particle is given by<span>\u00a0<\/span>Equation 2.53<span>\u00a0<\/span>and the kinetic energy is<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-21-Frame\"><span class=\"MathJax_MathContainer\"><span>K=E\u2212E0=(\u03b3\u22121)E0.<\/span><\/span><\/span><span>\u00a0<\/span>When the kinetic energy is known, we can invert<span>\u00a0<\/span>Equation 2.18<span>\u00a0<\/span>to find the momentum<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-22-Frame\"><span class=\"MathJax_MathContainer\"><span>p=(E2\u2212E02)\/c2=K(K+2E0)\/c<\/span><\/span><\/span><span>\u00a0<\/span>and substitute in<span>\u00a0<\/span>Equation 2.57<span>\u00a0<\/span>to obtain<\/p>\n<div data-type=\"equation\" id=\"fs-id1163712341589\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-23-Frame\">\n<div class=\"textbox\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-23-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb=hp=hcK(K+2E0).<\/span><\/span><\/div>\n<div class=\"os-equation-number\">\n<p><span class=\"os-number\">[2.58]<\/span><\/p>\n<\/div>\n<\/div>\n<p><span style=\"text-indent: 1em;font-size: 1rem\">Depending on the problem at hand, in this equation we can use the following values for<\/span><span style=\"text-indent: 1em;font-size: 1rem\">\u00a0<\/span><em style=\"text-indent: 1em;font-size: 1rem\" data-effect=\"italics\">hc<\/em><span style=\"text-indent: 1em;font-size: 1rem\">:<\/span><span style=\"text-indent: 1em;font-size: 1rem\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-24-Frame\" style=\"text-indent: 1em;font-size: 1rem\"><span class=\"MathJax_MathContainer\">hc=(6.626\u00d710\u221234J\u00b7s)(2.998\u00d7108m\/s)=1.986\u00d710\u221225J\u00b7m=1.241eV\u00b7\u03bcm<\/span><\/span><\/p>\n<\/div>\n<div class=\"os-equation-number\">\n<p><strong><span style=\"text-indent: 1em;font-size: 1rem\">Solution<\/span><\/strong><\/p>\n<\/div>\n<\/div>\n<ol id=\"fs-id1172101985706\" type=\"a\">\n<li>For the basketball, the kinetic energy is<br \/>\n<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163711937451\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-25-Frame\"><span class=\"MathJax_MathContainer\"><span>K=m0u2\/2=(0.65kg)(10m\/s)2\/2=32.5J<\/span><\/span><\/div>\n<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span>and the rest mass energy is<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163711938219\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-26-Frame\"><span class=\"MathJax_MathContainer\"><span>E0=m0c2=(0.65kg)(2.998\u00d7108m\/s)2=5.84\u00d71016J.<\/span><\/span><\/div>\n<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span>We see that<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-27-Frame\"><span class=\"MathJax_MathContainer\"><span>K\/(K+E0)\u226a1and usep=m0u=(0.65kg)(10m\/s)=6.5J\u00b7s\/m:<\/span><\/span><\/span><span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163710827302\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-28-Frame\">\n<p><span class=\"MathJax_MathContainer\"><span>\u03bb=hp=6.626\u00d710\u221234J\u00b7s6.5J\u00b7s\/m=1.02\u00d710\u221234m.<\/span><\/span><\/p>\n<\/div>\n<\/div>\n<\/li>\n<li>For the nonrelativistic electron,<br \/>\n<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163712317187\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-29-Frame\"><span class=\"MathJax_MathContainer\"><span>E0=m0c2=(9.109\u00d710\u221231kg)(2.998\u00d7108m\/s)2=511keV<\/span><\/span><\/div>\n<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span>and when<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-30-Frame\"><span class=\"MathJax_MathContainer\"><span>K=1.0eV,<\/span><\/span><\/span><span>\u00a0<\/span>we have<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-31-Frame\"><span class=\"MathJax_MathContainer\"><span>K\/(K+E0)=(1\/512)\u00d710\u22123\u226a1,<\/span><\/span><\/span><span>\u00a0<\/span>so we can use the nonrelativistic formula. However, it is simpler here to use<span>\u00a0<\/span>Equation 2.58:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163712141043\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-32-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb=hp=hcK(K+2E0)=1.241eV\u00b7\u03bcm(1.0eV)[1.0eV+2(511keV)]=1.23nm.<\/span><\/span><\/div>\n<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span>If we use nonrelativistic momentum, we obtain the same result because 1 eV is much smaller than the rest mass of the electron.<\/li>\n<li>For a fast electron with<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-33-Frame\"><span class=\"MathJax_MathContainer\"><span>K=108keV,<\/span><\/span><\/span><span>\u00a0<\/span>relativistic effects cannot be neglected because its total energy is<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-34-Frame\"><span class=\"MathJax_MathContainer\"><span>E=K+E0=108keV+511keV=619keV<\/span><\/span><\/span><span>\u00a0<\/span>and<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-35-Frame\"><span class=\"MathJax_MathContainer\"><span>K\/E=108\/619<\/span><\/span><\/span><span>\u00a0<\/span>is not negligible:<br \/>\n<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163712383831\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-36-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb=hp=hcK(K+2E0)=1.241eV\u00b7\u03bcm108keV[108keV+2(511keV)]=3.55pm.<\/span><\/span><\/div>\n<div><\/div>\n<div class=\"MathJax_MathML\">\n<p><span data-type=\"title\" style=\"text-indent: 1em;font-size: 1rem\"><strong>Significance<\/strong><\/span><\/p>\n<p><span style=\"text-indent: 1em;font-size: 1rem\">We see from these estimates that De Broglie\u2019s wavelengths of macroscopic objects such as a ball are immeasurably small. Therefore, even if they exist, they are not detectable and do not affect the motion of macroscopic objects.<\/span><\/p>\n<\/div>\n<\/div>\n<\/li>\n<\/ol>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<div class=\"check-understanding ui-has-child-title\" data-type=\"note\" id=\"fs-id1163712281992\">\n<header><\/header>\n<section>\n<div class=\"os-note-body\">\n<div class=\"os-hasSolution unnumbered\" data-type=\"exercise\" id=\"fs-id1163711954901\">\n<section>\n<div data-type=\"problem\" id=\"fs-id1163712212556\"><\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<div class=\"textbox textbox--key-takeaways\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\"><span class=\"os-title-label\">CHECK YOUR UNDERSTANDING<span>\u00a02<\/span><\/span><span class=\"os-number\">.11<\/span><\/p>\n<\/header>\n<div class=\"textbox__content\">\n<header><span style=\"font-size: 1rem\">What is de Broglie\u2019s wavelength of a nonrelativistic proton with a kinetic energy of 1.0 eV?<\/span><\/header>\n<\/div>\n<\/div>\n<p id=\"fs-id1163711963974\">Using the concept of the electron matter wave, de Broglie provided a rationale for the quantization of the electron\u2019s angular momentum in the hydrogen atom, which was postulated in Bohr\u2019s quantum theory. The physical explanation for the first Bohr quantization condition comes naturally when we assume that an electron in a hydrogen atom behaves not like a particle but like a wave. To see it clearly, imagine a stretched guitar string that is clamped at both ends and vibrates in one of its normal modes. If the length of the string is<span>\u00a0<\/span><em data-effect=\"italics\">l\u00a0<\/em>(Figure 2.18), the wavelengths of these vibrations cannot be arbitrary but must be such that an integer<span>\u00a0<\/span><em data-effect=\"italics\">k<\/em><span>\u00a0<\/span>number of half-wavelengths<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-37-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb\/2<\/span><\/span><\/span>fit exactly on the distance<span>\u00a0<\/span><em data-effect=\"italics\">l<\/em><span>\u00a0<\/span>between the ends. This is the condition<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-38-Frame\"><span class=\"MathJax_MathContainer\"><span>l=k\u03bb\/2<\/span><\/span><\/span><span>\u00a0<\/span>for a standing wave on a string. Now suppose that instead of having the string clamped at the walls, we bend its length into a circle and fasten its ends to each other. This produces a circular string that vibrates in normal modes, satisfying the same standing-wave condition, but the number of half-wavelengths must now be an even number<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-39-Frame\"><span class=\"MathJax_MathContainer\"><span>k,k=2n,<\/span><\/span><\/span><span>\u00a0<\/span>and the length<span>\u00a0<\/span><em data-effect=\"italics\">l<\/em><span>\u00a0<\/span>is now connected to the radius<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-40-Frame\"><span class=\"MathJax_MathContainer\"><span>rn<\/span><\/span><\/span><span>\u00a0<\/span>of the circle. This means that the radii are not arbitrary but must satisfy the following standing-wave condition:<\/p>\n<div data-type=\"equation\" id=\"fs-id1163711989563\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-41-Frame\">\n<div class=\"textbox\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-41-Frame\"><span class=\"MathJax_MathContainer\"><span>2\u03c0rn=2n\u03bb2.<\/span><\/span><\/div>\n<div class=\"os-equation-number\"><span class=\"os-number\">[2.59]<\/span><\/div>\n<\/div>\n<p><span style=\"font-size: 14pt\">If an electron in the<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><em style=\"font-size: 14pt\" data-effect=\"italics\">n<\/em><span style=\"font-size: 14pt\">th Bohr orbit moves as a wave, by<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">Equation 2.59<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">its wavelength must be equal to<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-42-Frame\" style=\"font-size: 14pt\"><span class=\"MathJax_MathContainer\">\u03bb=2\u03c0rn\/n.<\/span><\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">Assuming that<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">Equation 2.58<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">is valid, the electron wave of this wavelength corresponds to the electron\u2019s linear momentum,<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-43-Frame\" style=\"font-size: 14pt\"><span class=\"MathJax_MathContainer\">p=h\/\u03bb=nh\/(2\u03c0rn)=n\u210f\/rn.<\/span><\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">In a circular orbit, therefore, the electron\u2019s angular momentum must be<\/span><\/p>\n<\/div>\n<\/div>\n<div data-type=\"equation\" id=\"fs-id1163712425771\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-44-Frame\">\n<div class=\"textbox\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-44-Frame\"><span class=\"MathJax_MathContainer\"><span>Ln=rnp=rnn\u210frn=n\u210f.<\/span><\/span><\/div>\n<div class=\"os-equation-number\"><span class=\"os-number\">[2.60]<\/span><\/div>\n<\/div>\n<p><span style=\"font-size: 14pt\">This equation is the first of Bohr\u2019s quantization conditions, given by<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">Equation 2.36. Providing a physical explanation for Bohr\u2019s quantization condition is a convincing theoretical argument for the existence of matter waves.<\/span><\/p>\n<\/div>\n<\/div>\n<div class=\"os-figure\">\n<figure id=\"CNX_UPhysics_39_05_string\">\n<figure style=\"width: 996px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" alt=\"Figure A is the standing-wave pattern for a string clamped in the wall. The distance between each node corresponds to the half gamma. Figure B is the standing-wave pattern for an electron wave trapped in the third Bohr orbit in the hydrogen atom. The wave has a circular shape with six nodes. The distance between each two node corresponds to the gamma.\" data-media-type=\"image\/jpeg\" id=\"27810\" src=\"https:\/\/cnx.org\/resources\/9959a8324ba37cdbf28a48d785a8d8787b4669b6\" width=\"996\" height=\"295\" \/><figcaption class=\"wp-caption-text\">Figure 2.18 Standing-wave pattern: (a) a stretched string clamped at the walls; (b) an electron wave trapped in the third Bohr orbit in the hydrogen atom.<\/figcaption><\/figure>\n<\/figure>\n<\/div>\n<div data-type=\"example\" id=\"fs-id1163710819156\" class=\"ui-has-child-title\">\n<header><\/header>\n<section>\n<div class=\"textbox shaded\">\n<header>\n<h3 class=\"os-title\"><span class=\"os-title-label\">EXAMPLE<span>\u00a02<\/span><\/span><span class=\"os-number\">.12<\/span><span class=\"os-divider\"><\/span><\/h3>\n<\/header>\n<section>\n<p id=\"fs-id1172099500868\"><span data-type=\"title\"><strong>The Electron Wave in the Ground State of Hydrogen<\/strong><\/span><\/p>\n<p>Find the de Broglie wavelength of an electron in the ground state of hydrogen.<\/p>\n<p><span data-type=\"title\"><strong>Strategy<\/strong><\/span><\/p>\n<p>We combine the first quantization condition in<span>\u00a0<\/span>Equation 2.60<span>\u00a0<\/span>with<span>\u00a0<\/span>Equation 2.36<span>\u00a0<\/span>and use<span>\u00a0<\/span>Equation 2.38<span>\u00a0<\/span>for the first Bohr radius with<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-45-Frame\"><span class=\"MathJax_MathContainer\"><span>n=1.<\/span><\/span><\/span><\/p>\n<p><span data-type=\"title\"><strong>Solution<\/strong><\/span><\/p>\n<p>When<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-46-Frame\"><span class=\"MathJax_MathContainer\"><span>n=1<\/span><\/span><\/span><span>\u00a0<\/span>and<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-47-Frame\"><span class=\"MathJax_MathContainer\"><span>rn=a0=0.529\u00c5,<\/span><\/span><\/span><span>\u00a0<\/span>the Bohr quantization condition gives<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-48-Frame\"><span class=\"MathJax_MathContainer\"><span>a0p=1\u00b7\u210f\u21d2p=\u210f\/a0.<\/span><\/span><\/span><span>\u00a0<\/span>The electron wavelength is:<\/p>\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163710983371\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-49-Frame\">\n<p><span class=\"MathJax_MathContainer\"><span>\u03bb=h\/p=h\/\u210f\/a0=2\u03c0a0=2\u03c0(0.529\u00c5)=3.324\u00c5.<\/span><\/span><\/p>\n<p><span data-type=\"title\" style=\"text-indent: 1em;font-size: 1rem\"><strong>Significance<\/strong><\/span><\/p>\n<p><span style=\"text-indent: 1em;font-size: 1rem\">We obtain the same result when we use<\/span><span style=\"text-indent: 1em;font-size: 1rem\">\u00a0<\/span>Equation 2.58<span style=\"text-indent: 1em;font-size: 1rem\">\u00a0<\/span><span style=\"text-indent: 1em;font-size: 1rem\">directly.<\/span><\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<div class=\"textbox textbox--key-takeaways\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\"><span class=\"os-title-label\">CHECK YOUR UNDERSTANDING<span>\u00a02<\/span><\/span><span class=\"os-number\">.12<\/span><\/p>\n<\/header>\n<div class=\"textbox__content\">\n<header><span style=\"font-size: 1rem\">Find the de Broglie wavelength of an electron in the third excited state of hydrogen.<\/span><\/header>\n<\/div>\n<\/div>\n<p id=\"fs-id1163712285476\">Experimental confirmation of matter waves came in 1927 when C.<span>\u00a0<\/span><span class=\"no-emphasis\" data-type=\"term\" id=\"term262\">Davisson<\/span><span>\u00a0<\/span>and L.<span>\u00a0<\/span><span class=\"no-emphasis\" data-type=\"term\" id=\"term263\">Germer<\/span><span>\u00a0<\/span>performed a series of electron-scattering experiments that clearly showed that electrons do behave like waves. Davisson and Germer did not set up their experiment to confirm de Broglie\u2019s hypothesis: The confirmation came as a byproduct of their routine experimental studies of metal surfaces under electron bombardment.<\/p>\n<p id=\"fs-id1163712032514\">In the particular experiment that provided the very first evidence of electron waves (known today as the<span>\u00a0<\/span><span data-type=\"term\" id=\"term264\">Davisson\u2013Germer experiment<\/span>), they studied a surface of nickel. Their nickel sample was specially prepared in a high-temperature oven to change its usual polycrystalline structure to a form in which large single-crystal domains occupy the volume.<span>\u00a0<\/span>Figure 2.19<span>\u00a0<\/span>shows the experimental setup. Thermal electrons are released from a heated element (usually made of tungsten) in the electron gun and accelerated through a potential difference<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-50-Frame\"><span class=\"MathJax_MathContainer\"><span>\u0394V,<\/span><\/span><\/span><span>\u00a0<\/span>becoming a well-collimated beam of electrons produced by an electron gun. The kinetic energy<span>\u00a0<\/span><em data-effect=\"italics\">K<\/em><span>\u00a0<\/span>of the electrons is adjusted by selecting a value of the potential difference in the electron gun. This produces a beam of electrons with a set value of linear momentum, in accordance with the conservation of energy:<\/p>\n<div data-type=\"equation\" id=\"fs-id1163711981923\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-51-Frame\">\n<div class=\"textbox\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-51-Frame\"><span class=\"MathJax_MathContainer\"><span>e\u0394V=K=p22m\u21d2p=2me\u0394V.<\/span><\/span><\/div>\n<div class=\"os-equation-number\"><span class=\"os-number\">[2.61]<\/span><\/div>\n<\/div>\n<p><span style=\"font-size: 14pt\">The electron beam is incident on the nickel sample in the direction normal to its surface. At the surface, it scatters in various directions. The intensity of the beam scattered in a selected direction<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-52-Frame\" style=\"font-size: 14pt\"><span class=\"MathJax_MathContainer\">\u03c6<\/span><\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">is measured by a highly sensitive detector. The detector\u2019s angular position with respect to the direction of the incident beam can be varied from<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-53-Frame\" style=\"font-size: 14pt\"><span class=\"MathJax_MathContainer\">\u03c6=0\u00b0<\/span><\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">to<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-54-Frame\" style=\"font-size: 14pt\"><span class=\"MathJax_MathContainer\">\u03c6=90\u00b0.<\/span><\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">The entire setup is enclosed in a vacuum chamber to prevent electron collisions with air molecules, as such thermal collisions would change the electrons\u2019 kinetic energy and are not desirable.<\/span><\/p>\n<\/div>\n<\/div>\n<div class=\"os-figure\">\n<figure id=\"CNX_UPhysics_39_05_davisson\">\n<figure style=\"width: 457px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" alt=\"Figure shows the schematics of the experimental setup of the Davisson\u2013Germer diffraction experiment. A beam of electrons is emitted by the electron gun, passes through the collimator, and hits Nickel target. Diffracted beam forms an angle phi with the incident beam and is detected by a moving detector. All of this is shown happening in a vacuum\" data-media-type=\"image\/jpeg\" id=\"48111\" src=\"https:\/\/cnx.org\/resources\/8e4ffc6549ef9a8e810bde4b4ae984c3085a89d9\" width=\"457\" height=\"336\" \/><figcaption class=\"wp-caption-text\">Figure 2.19 Schematics of the experimental setup of the Davisson\u2013Germer diffraction experiment. A well-collimated beam of electrons is scattered off the nickel target. The kinetic energy of electrons in the incident beam is selected by adjusting a variable potential, \u0394V, in the electron gun. Intensity of the scattered electron beam is measured for a range of scattering angles \u03c6, whereas the distance between the detector and the target does not change.<\/figcaption><\/figure>\n<\/figure>\n<\/div>\n<p id=\"fs-id1163711938476\">When the nickel target has a polycrystalline form with many randomly oriented microscopic crystals, the incident electrons scatter off its surface in various random directions. As a result, the intensity of the scattered electron beam is much the same in any direction, resembling a diffuse reflection of light from a porous surface. However, when the nickel target has a regular crystalline structure, the intensity of the scattered electron beam shows a clear maximum at a specific angle and the results show a clear diffraction pattern (see<span>\u00a0<\/span>Figure 2.20). Similar diffraction patterns formed by X-rays scattered by various crystalline solids were studied in 1912 by father-and-son physicists William H.<span>\u00a0<\/span><span class=\"no-emphasis\" data-type=\"term\" id=\"term265\">Bragg<\/span><span>\u00a0<\/span>and William L.<span>\u00a0<\/span><span class=\"no-emphasis\" data-type=\"term\" id=\"term266\">Bragg<\/span>. The Bragg law in X-ray crystallography provides a connection between the wavelength<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-57-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb<\/span><\/span><\/span><span>\u00a0<\/span>of the radiation incident on a crystalline lattice, the lattice spacing, and the position of the interference maximum in the diffracted radiation (see<span>\u00a0<\/span><a class=\"target-chapter\" href=\"https:\/\/cnx.org\/contents\/af275420-6050-4707-995c-57b9cc13c358@11.1:530d811a-3df9-4913-9c8b-5844eae74384@5\" data-page=\"28\">Diffraction<\/a>).<\/p>\n<p id=\"fs-id1163710853034\">The lattice spacing of the Davisson\u2013Germer target, determined with X-ray crystallography, was measured to be<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-58-Frame\"><span class=\"MathJax_MathContainer\"><span>a=2.15\u00c5.<\/span><\/span><\/span><span>\u00a0<\/span>Unlike X-ray crystallography in which X-rays penetrate the sample, in the original Davisson\u2013Germer experiment, only the surface atoms interact with the incident electron beam. For the surface diffraction, the maximum intensity of the reflected electron beam is observed for scattering angles that satisfy the condition<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-59-Frame\"><span class=\"MathJax_MathContainer\"><span>n\u03bb=asin\u03c6<\/span><\/span><\/span><span>\u00a0<\/span>(see<span>\u00a0<\/span>Figure 2.21). The first-order maximum (for<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-60-Frame\"><span class=\"MathJax_MathContainer\"><span>n=1<\/span><\/span><\/span>) is measured at a scattering angle of<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-61-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03c6\u224850\u00b0<\/span><\/span><\/span><span>\u00a0<\/span>at<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-62-Frame\"><span class=\"MathJax_MathContainer\"><span>\u0394V\u224854V,<\/span><\/span><\/span><span>\u00a0<\/span>which gives the wavelength of the incident radiation as<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-63-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb=(2.15\u00c5)sin50\u00b0=1.64\u00c5.<\/span><\/span><\/span><span>\u00a0<\/span>On the other hand, a 54-V potential accelerates the incident electrons to kinetic energies of<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-64-Frame\"><span class=\"MathJax_MathContainer\"><span>K=54eV.<\/span><\/span><\/span><span>\u00a0<\/span>Their momentum, calculated from<span>\u00a0<\/span>Equation 2.61, is<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-65-Frame\"><span class=\"MathJax_MathContainer\"><span>p=2.478\u00d710\u22125eV\u00b7s\/m.<\/span><\/span><\/span><span>\u00a0<\/span>When we substitute this result in<span>\u00a0<\/span>Equation 2.58, the de Broglie wavelength is obtained as<\/p>\n<div data-type=\"equation\" id=\"fs-id1163712085202\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-66-Frame\">\n<div class=\"textbox\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-66-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb=hp=4.136\u00d710\u221215eV\u00b7s2.478\u00d710\u22125eV\u00b7s\/m=1.67\u00c5.<\/span><\/span><\/div>\n<div class=\"os-equation-number\"><span class=\"os-number\">[2.62]<\/span><\/div>\n<\/div>\n<p><span style=\"font-size: 14pt\">The same result is obtained when we use<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-67-Frame\" style=\"font-size: 14pt\"><span class=\"MathJax_MathContainer\">K=54eV<\/span><\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">in<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">Equation 2.61. The proximity of this theoretical result to the Davisson\u2013Germer experimental value of<\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-68-Frame\" style=\"font-size: 14pt\"><span class=\"MathJax_MathContainer\">\u03bb=1.64\u00c5<\/span><\/span><span style=\"font-size: 14pt\">\u00a0<\/span><span style=\"font-size: 14pt\">is a convincing argument for the existence of de Broglie matter waves.<\/span><\/p>\n<\/div>\n<\/div>\n<div class=\"os-figure\">\n<figure id=\"CNX_UPhysics_39_05_davisson2\">\n<figure style=\"width: 347px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" alt=\"The graph shows the dependence of the intensity of the scattering beam on the scattering angle in degrees. The intensity degrees from 10 to 30 degrees, followed by a sharp increase and maximum at 50 degrees, and then reaches zero at 80 degrees.\" data-media-type=\"image\/jpeg\" id=\"21884\" src=\"https:\/\/cnx.org\/resources\/822ba59021544c10997c333c09a9007ff22072b5\" width=\"347\" height=\"346\" \/><figcaption class=\"wp-caption-text\">Figure 2.20 The experimental results of electron diffraction on a nickel target for the accelerating potential in the electron gun of about \u0394V=54V: The intensity maximum is registered at the scattering angle of about \u03c6=50\u00b0.<\/figcaption><\/figure>\n<p>&nbsp;<\/figure>\n<\/div>\n<div class=\"os-figure\">\n<figure id=\"CNX_UPhysics_39_05_bragg\">\n<figure style=\"width: 333px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" alt=\"Figure shows the surface diffraction of a monochromatic electromagnetic wave on a crystalline lattice structure. The in-phase incident beams are reflected from atoms on the surface. Phi is the angle between the incident and the reflected beam, the in-plane distance between the atoms is a.\" data-media-type=\"image\/jpeg\" id=\"28527\" src=\"https:\/\/cnx.org\/resources\/b9cd5291754f7d099d4511e8045535dd97197ce6\" width=\"333\" height=\"249\" \/><figcaption class=\"wp-caption-text\">Figure 2.21 In the surface diffraction of a monochromatic electromagnetic wave on a crystalline lattice structure, the in-phase incident beams are reflected from atoms on the surface. A ray reflected from the left atom travels an additional distance D=asin\u03c6 to the detector, where a is the lattice spacing. The reflected beams remain in-phase when D is an integer multiple of their wavelength \u03bb. The intensity of the reflected waves has pronounced maxima for angles \u03c6 satisfying n\u03bb=asin\u03c6.<\/figcaption><\/figure>\n<\/figure>\n<\/div>\n<p id=\"fs-id1163711996022\">Diffraction lines measured with low-energy electrons, such as those used in the Davisson\u2013Germer experiment, are quite broad (see<span>\u00a0<\/span>Figure 2.20) because the incident electrons are scattered only from the surface. The resolution of diffraction images greatly improves when a higher-energy electron beam passes through a thin metal foil. This occurs because the diffraction image is created by scattering off many crystalline planes inside the volume, and the maxima produced in scattering at Bragg angles are sharp (see<span>\u00a0<\/span>Figure 2.22).<\/p>\n<div class=\"os-figure\">\n<figure id=\"CNX_UPhysics_39_05_laue\">\n<figure style=\"width: 975px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" alt=\"Picture A is a photograph of the diffraction pattern obtained in scattering on a crystalline solid with X-rays. Picture B is a photograph of the diffraction pattern obtained in scattering on a crystalline solid with electrons. Both pictures demonstrate diffracted spots symmetrically arranged around the central beam.\" data-media-type=\"image\/jpeg\" id=\"13095\" src=\"https:\/\/cnx.org\/resources\/b261ff9f58dc5c1f3adbd87b482fbeb0839e5853\" width=\"975\" height=\"522\" \/><figcaption class=\"wp-caption-text\">Figure 2.22 Diffraction patterns obtained in scattering on a crystalline solid: (a) with X-rays, and (b) with electrons. The observed pattern reflects the symmetry of the crystalline structure of the sample.<span class=\"os-caption\"><\/span><\/figcaption><\/figure>\n<\/figure>\n<\/div>\n<p id=\"fs-id1163711912089\">Since the work of Davisson and Germer, de Broglie\u2019s hypothesis has been extensively tested with various experimental techniques, and the existence of de Broglie waves has been confirmed for numerous elementary particles. Neutrons have been used in scattering experiments to determine crystalline structures of solids from interference patterns formed by neutron matter waves. The neutron has zero charge and its mass is comparable with the mass of a positively charged proton. Both neutrons and protons can be seen as matter waves. Therefore, the property of being a matter wave is not specific to electrically charged particles but is true of all particles in motion. Matter waves of molecules as large as carbon<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-75-Frame\"><span class=\"MathJax_MathContainer\"><span>C60<\/span><\/span><\/span><span>\u00a0<\/span>have been measured. All physical objects, small or large, have an associated matter wave as long as they remain in motion. The universal character of de Broglie matter waves is firmly established.<\/p>\n<div data-type=\"example\" id=\"fs-id1163712192182\" class=\"ui-has-child-title\">\n<header><\/header>\n<section>\n<div class=\"textbox shaded\">\n<header>\n<h3 class=\"os-title\"><span class=\"os-title-label\">EXAMPLE<span>\u00a02<\/span><\/span><span class=\"os-number\">.13<\/span><span class=\"os-divider\"><\/span><\/h3>\n<\/header>\n<section>\n<p id=\"fs-id1172102099642\"><span data-type=\"title\"><strong>Neutron Scattering<\/strong><\/span><\/p>\n<p>Suppose that a neutron beam is used in a diffraction experiment on a typical crystalline solid. Estimate the kinetic energy of a neutron (in eV) in the neutron beam and compare it with kinetic energy of an ideal gas in equilibrium at room temperature.<\/p>\n<p><span data-type=\"title\"><strong>Strategy<\/strong><\/span><\/p>\n<p>We assume that a typical crystal spacing<span>\u00a0<\/span><em data-effect=\"italics\">a<\/em><span>\u00a0<\/span>is of the order of 1.0 \u00c5. To observe a diffraction pattern on such a lattice, the neutron wavelength<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-76-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb<\/span><\/span><\/span><span>\u00a0<\/span>must be on the same order of magnitude as the lattice spacing. We use<span>\u00a0<\/span>Equation 2.61<span>\u00a0<\/span>to find the momentum<span>\u00a0<\/span><em data-effect=\"italics\">p<\/em><span>\u00a0<\/span>and kinetic energy<span>\u00a0<\/span><em data-effect=\"italics\">K<\/em>. To compare this energy with the energy<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-77-Frame\"><span class=\"MathJax_MathContainer\"><span>ET<\/span><\/span><\/span><span>\u00a0<\/span>of ideal gas in equilibrium at room temperature<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-78-Frame\"><span class=\"MathJax_MathContainer\"><span>T=300K,<\/span><\/span><\/span><span>\u00a0<\/span>we use the relation<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-79-Frame\"><span class=\"MathJax_MathContainer\"><span>K=32kBT,<\/span><\/span><\/span><span>\u00a0<\/span>where<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-80-Frame\"><span class=\"MathJax_MathContainer\"><span>kB=8.62\u00d710\u22125eV\/K<\/span><\/span><\/span><span>\u00a0<\/span>is the Boltzmann constant.<\/p>\n<p><span data-type=\"title\"><strong>Solution<\/strong><\/span><\/p>\n<p>We evaluate<span>\u00a0<\/span><em data-effect=\"italics\">pc<\/em><span>\u00a0<\/span>to compare it with the neutron\u2019s rest mass energy<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-81-Frame\"><span class=\"MathJax_MathContainer\"><span>E0=940MeV:<\/span><\/span><\/span><\/p>\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163711912453\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-82-Frame\">\n<p><span class=\"MathJax_MathContainer\"><span>p=h\u03bb\u21d2pc=hc\u03bb=1.241\u00d710\u22126eV\u00b7m10\u221210m=12.41keV.<\/span><\/span><\/p>\n<p><span style=\"text-indent: 1em;font-size: 1rem\">We see that<\/span><span style=\"text-indent: 1em;font-size: 1rem\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-83-Frame\" style=\"text-indent: 1em;font-size: 1rem\"><span class=\"MathJax_MathContainer\">p2c2\u226aE02<\/span><\/span><span style=\"text-indent: 1em;font-size: 1rem\">\u00a0<\/span><span style=\"text-indent: 1em;font-size: 1rem\">so<\/span><span style=\"text-indent: 1em;font-size: 1rem\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-84-Frame\" style=\"text-indent: 1em;font-size: 1rem\"><span class=\"MathJax_MathContainer\">K\u226aE0<\/span><\/span><span style=\"text-indent: 1em;font-size: 1rem\">\u00a0<\/span><span style=\"text-indent: 1em;font-size: 1rem\">and we can use the nonrelativistic kinetic energy:<\/span><\/p>\n<p><span style=\"font-size: 1rem;text-indent: 0px\">K=p22mn=h22\u03bb2mn=(6.63\u00d710\u221234J\u00b7s)2(2\u00d710\u221220m2)(1.66\u00d710\u221227kg)=1.32\u00d710\u221220J=82.7meV.<\/span><\/p>\n<\/div>\n<\/div>\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163711892811\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-85-Frame\">\n<p><span style=\"text-indent: 1em;font-size: 1rem\">Kinetic energy of ideal gas in equilibrium at 300 K is:<\/span><\/p>\n<\/div>\n<\/div>\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163712283018\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-86-Frame\">\n<p><span class=\"MathJax_MathContainer\"><span class=\"MathJax_MathContainer\"><span>KT=32kBT=32(8.62\u00d710\u22125eV\/K)(300K)=38.8MeV.<\/span><\/span><\/span><span style=\"text-indent: 1em;font-size: 1rem\"><span style=\"text-indent: 1em;font-size: 1rem\"><span style=\"text-indent: 1em;font-size: 1rem\">We see that these energies are of the same order of magnitude.<\/span><\/span><\/span><\/p>\n<p><span data-type=\"title\" style=\"text-indent: 1em;font-size: 1rem\"><span data-type=\"title\" style=\"text-indent: 1em;font-size: 1rem\"><strong>Significance<\/strong><\/span><\/span><\/p>\n<p><span style=\"font-size: 1rem;text-indent: 1em\">Neutrons with energies in this range, which is typical for an ideal gas at room temperature, are called \u201cthermal neutrons.\u201d<\/span><\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<div class=\"textbox shaded\">\n<header>\n<h3 class=\"os-title\"><span class=\"os-title-label\">EXAMPLE<span>\u00a02<\/span><\/span><span class=\"os-number\">.14<\/span><span class=\"os-divider\"><\/span><\/h3>\n<\/header>\n<section>\n<p id=\"fs-id1172099743389\"><span data-type=\"title\"><strong>Wavelength of a Relativistic Proton<\/strong><\/span><\/p>\n<p>In a supercollider at CERN, protons can be accelerated to velocities of 0.75<em data-effect=\"italics\">c<\/em>. What are their de Broglie wavelengths at this speed? What are their kinetic energies?<\/p>\n<p><span data-type=\"title\"><strong>Strategy<\/strong><\/span><\/p>\n<p>The rest mass energy of a proton is<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-87-Frame\"><span class=\"MathJax_MathContainer\"><span>E0=m0c2=(1.672\u00d710\u221227kg)(2.998\u00d7108m\/s)2=938MeV.<\/span><\/span><\/span><span>\u00a0<\/span>When the proton\u2019s velocity is known, we have<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-88-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03b2=0.75<\/span><\/span><\/span><span>\u00a0<\/span>and<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-89-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03b2\u03b3=0.75\/1\u22120.752=1.714.<\/span><\/span><\/span><span>\u00a0<\/span>We obtain the wavelength<span>\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-90-Frame\"><span class=\"MathJax_MathContainer\"><span>\u03bb<\/span><\/span><\/span><span>\u00a0<\/span>and kinetic energy<span>\u00a0<\/span><em data-effect=\"italics\">K<\/em><span>\u00a0<\/span>from relativistic relations.<\/p>\n<p><strong>Solution<\/strong><\/p>\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163711940582\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-91-Frame\">\n<p><span class=\"MathJax_MathContainer\"><span>\u03bb=hp=hcpc=hc\u03b2\u03b3E0=1.241eV\u00b7\u03bcm1.714(938MeV)=0.77fm<\/span><\/span><\/p>\n<p><span style=\"font-size: 1rem;text-indent: 0px\">K=E0(\u03b3\u22121)=938MeV(1\/1\u22120.752\u22121)=480.1MeV<\/span><\/p>\n<\/div>\n<\/div>\n<div class=\"unnumbered\" data-label=\"\" data-type=\"equation\" id=\"fs-id1163712167572\">\n<div class=\"MathJax_MathML\" id=\"MathJax-Element-92-Frame\">\n<p><span data-type=\"title\" style=\"text-indent: 1em;font-size: 1rem\"><strong>Significance<\/strong><\/span><\/p>\n<p><span style=\"text-indent: 1em;font-size: 1rem\">Notice that because a proton is 1835 times more massive than an electron, if this experiment were performed with electrons, a simple rescaling of these results would give us the electron\u2019s wavelength of<\/span><span style=\"text-indent: 1em;font-size: 1rem\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-93-Frame\" style=\"text-indent: 1em;font-size: 1rem\"><span class=\"MathJax_MathContainer\">(1835)0.77fm=1.4pm<\/span><\/span><span style=\"text-indent: 1em;font-size: 1rem\">\u00a0<\/span><span style=\"text-indent: 1em;font-size: 1rem\">and its kinetic energy of<\/span><span style=\"text-indent: 1em;font-size: 1rem\">\u00a0<\/span><span class=\"MathJax_MathML\" id=\"MathJax-Element-94-Frame\" style=\"text-indent: 1em;font-size: 1rem\"><span class=\"MathJax_MathContainer\">480.1MeV\/1835=261.6keV.<\/span><\/span><\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<div class=\"textbox textbox--key-takeaways\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\"><span class=\"os-title-label\">CHECK YOUR UNDERSTANDING<span>\u00a02<\/span><\/span><span class=\"os-number\">.13<\/span><\/p>\n<\/header>\n<div class=\"textbox__content\">\n<header>\n<div class=\"os-title\"><span style=\"font-size: 1rem\">Find the de Broglie wavelength and kinetic energy of a free electron that travels at a speed of 0.75<\/span><em style=\"font-size: 1rem\" data-effect=\"italics\">c<\/em><span style=\"font-size: 1rem\">.<\/span><\/div>\n<\/header>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<div class=\"textbox\"><em>Download for free at http:\/\/cnx.org\/contents\/af275420-6050-4707-995c-57b9cc13c358@11.1<\/em><\/div>\n","protected":false},"author":615,"menu_order":6,"template":"","meta":{"pb_show_title":"on","pb_short_title":"2. Photons and Matter Waves","pb_subtitle":"2.5 De Broglie\u2019s Matter Waves","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-164","chapter","type-chapter","status-publish","hentry"],"part":36,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/bcitphys8400\/wp-json\/pressbooks\/v2\/chapters\/164","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/bcitphys8400\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/bcitphys8400\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/bcitphys8400\/wp-json\/wp\/v2\/users\/615"}],"version-history":[{"count":8,"href":"https:\/\/pressbooks.bccampus.ca\/bcitphys8400\/wp-json\/pressbooks\/v2\/chapters\/164\/revisions"}],"predecessor-version":[{"id":439,"href":"https:\/\/pressbooks.bccampus.ca\/bcitphys8400\/wp-json\/pressbooks\/v2\/chapters\/164\/revisions\/439"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/bcitphys8400\/wp-json\/pressbooks\/v2\/parts\/36"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/bcitphys8400\/wp-json\/pressbooks\/v2\/chapters\/164\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/bcitphys8400\/wp-json\/wp\/v2\/media?parent=164"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/bcitphys8400\/wp-json\/pressbooks\/v2\/chapter-type?post=164"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/bcitphys8400\/wp-json\/wp\/v2\/contributor?post=164"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/bcitphys8400\/wp-json\/wp\/v2\/license?post=164"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}