CHEMISTRY 161
HW#7A
32, 34, 38, 40, 44, 46, 50, 52, 54, 58, 60, 66.
7-32
A photon is the smallest indivisible packet or particle of light energy
7-34
If we simply apply the equation E = hc/, we definitely would think that the photon moving away
from us has lost energy. However, we have to consider here the frame of reference. It only appears to
us that the photon lost energy due to its motion relative to us. If we were traveling right along with the
photon the energy would not change. From the point of view of the photon, its energy has not
changed.
7-38
6.24  10 –19 J
 9.417  1014 s –1
6.626  10 –34 J  s
3.00  108 m/s

 3.19  10 –7 m or 319 nm
9.417  1014 s –1

7-40
The energy of light of 500 nm wavelength can be found as
6.626  10 –34 J  s  3.00  108 m/s
E
 3.98  10 –19 J
5.00  10 –7 m
This energy is less than the work function for tantalum ( = 6.81  10–19 J), so tantalum could not be
used to convert solar energy at 500 nm to electricity.
7-44
Energy of the photons of light at 500 nm:
6.626  10 –34 J  s  3.00  108 m/s
E
= 3.975  10 –19 J/photon
5.00  10 –7 m
Number of photons in 1 m3:
1  10 –15 J
1 photon

 2520 photons/m3
m3
3.975  10 –19 J
7-46
The principal quantum number, n, describes, in part, the energy of an electron and can be seen as a
description of how far away from the nucleus the electron resides (in an orbit).
7-50
Because energy levels in hydrogen become more closely spaced as n increases, ∆E for the energy
levels at lower n values must be greater and therefore have higher frequencies associated with them.
In order of increasing frequencies:
(d) n = 11 to n = 13 < (c) n = 9 to n = 11 < (b) n = 6 to n = 8 < (a) n = 4 to n = 6
7-52
These emissions are in the Lyman series which all fall into the ultraviolet (UV) region of the
spectrum.
7-54
Because both atoms have only one electron, there is no e –– e– repulsion in the atoms. Their spectra,
therefore, should both be fairly simple, as should emissions based on the transitions from higher to
lower n orbits. The He+ ion, however, pulls the electron in closer to the nucleus because of its
increased nuclear charge. This shifts all the n orbits to lower energy, and the ∆n energies are observed
at shorter wavelengths.
7-58
 1 1
E  (2.18  10–18 J)(2) 2  2 – 2   1.21  10–18 J
2 3 
6.626  10 –34 J  s  3.00  108 m/s
 1.64  10 –7 m or 164 nm
1.21  10 –18 J
This is not in the visible range. Therefore, the transition from higher energy states to n = 2 in He+ will
never produce visible light.

7-60
(a)
Because this line is probably in the Lyman series, the transition is associated with the
emission of light from an electron in an excited state relaxing to the ground state (n = 1).
(b) Here, ni = n2.
1 1
1
 [1.097  10 –2 (nm) –1 ]  2  2 
92.3 nm
1 n 
2
1 1

 0.988
12 n22
1
 0.012
n22
n2  9
(c) The longest wavelength (lowest energy) transition from the ground state (n = 1) is n = 1 to n = 2.
 1 1
E  2.18  10–18 J  2 – 2   1.64  10–18 J
2 1 
7-66
(a) False. Heavier particles have a shorter wavelength than lighter particles.
(b) True. When m2 = 2m1, then
 2 h 2m1u 1


1
h m1u
2
(c) True. Doubling the speed gives u2 = 2u1:
 2 h m2u1 1


1
h mu1
2
This is the same as doubling the mass in b.