Chapter 2: Photons and Matter Waves
Chapter 2 Answer Key
Chapter 2 Answer Key
Check Your Understanding
Bunsen’s burner
The wavelength of the radiation maximum decreases with increasing temperature.
Tα/Tβ=1/3≅0.58, so the star β is hotter.
3.3×10−19J
No, because then ΔE/E≈10−21
−0.91 V; 1040 nm
h=6.40×10−34J·s=4.0×10−15eV·s;−3.5%
(Δλ)min=0m at a 0° angle; 71.0pm+0.5λc=72.215pm
121.5 nm and 91.1 nm; no, these spectral bands are in the ultraviolet
v2=1.1×106m/s≅0.0036c; L2=2ℏK2=3.4eV
29 pm
λ=2πna0=2(3.324Å)=6.648Å
λ=2.14pm; K=261.56keV
0.052°
doubles it
Conceptual Questions
1.
yellow
3.
goes from red to violet through the rainbow of colors
5.
would not differ
7.
human eye does not see IR radiation
9.
No
11.
from the slope
13.
Answers may vary
15.
the particle character
17.
Answers may vary
19.
no; yes
21.
no
23.
right angle
25.
no
27.
They are at ground state.
29.
Answers may vary
31.
increase
33.
for larger n
35.
Yes, the excess of 13.6 eV will become kinetic energy of a free electron.
37.
no
39.
X-rays, best resolving power
41.
proton
43.
negligibly small de Broglie’s wavelengths
45.
to avoid collisions with air molecules
47.
Answers may vary
49.
Answers may vary
51.
yes
53.
yes
Problems
55.
a. 0.81 eV; b. 2.1×1023; c. 2 min 20 sec
57.
a. 7245 K; b. 3.62 μm
59.
about 3 K
61.
4.835×1018 Hz; 0.620 Å
63.
263 nm; no
65.
3.68 eV
67.
4.09 eV
69.
5.54 eV
71.
a. 1.89 eV; b. 459 THz; c. 1.21 V
73.
264 nm; UV
75.
1.95×106m/s
77.
1.66×10−32kg·m/s
79.
56.21 eV
81.
6.63×10−23kg·m/s; 124 keV
83.
82.9 fm; 15 MeV
85.
(Proof)
87.
Δλ30/Δλ45=45.74%
89.
121.5 nm
91.
a. 0.661 eV; b. –10.2 eV; c. 1.511 eV
93.
3038 THz
95.
97.33 nm
97.
a. h/π; b. 3.4 eV; c. – 6.8 eV; d. – 3.4 eV
99.
n=4
101.
365 nm; UV
103.
no
105.
7
107.
145.5 pm
109.
20 fm; 9 fm
111.
a. 2.103 eV; b. 0.846 nm
113.
80.9 pm
115.
2.21×10−19m/s
117.
9.929×1032
119.
γ=1060; 0.00124 fm
121.
24.11 V
123.
a. P=2I/c=8.67×10−6N/m2; b. a=PA/m=8.67×10-4m/s2; c. 74.91 m/s
125.
x=4.965
Additional Problems
127.
7.124×1016W/m3
129.
1.034 eV
131.
5.93×1018
133.
387.8 nm
135.
a. 4.02×1015; b. 0.533 mW
137.
a. 4.02×1015; b. 0.533 mW; c. 0.644 mA; d. 2.57 ns
139.
a. 0.132 pm; b. 9.39 MeV; c. 0.047 MeV
141.
a. 2 kJ; b. 1.33×10−5kg·m/s; c. 1.33×10−5N; d. yes
143.
a. 0.003 nm; b. 105.56°
145.
n=3
147.
a. a0/2; b. −54.4eV/n2; c. a0/3,−122.4eV/n2
149.
a. 36; b. 18.2 nm; c. UV
151.
396 nm; 5.23 neV
153.
7.3 keV
155.
728 m/s; 1.5μV
157.
λ=hc/K(2E0+K)=3.705×10−12m,K=100keV
159.
Δλc(electron)/Δλc(proton)=mp/me=1836
161.
(Proof)
163.
5.1×1017Hz