Answers to selected problems, Chapter 7
Review questions
7. Not only is there an obvious difference in the concept of ordering elements by mass versus by number of protons, but there is a practical difference, too: in the jump from Te to I, the number of protons increases by one but the total mass decreases. One exception like this is enough to break the rule.
22. Bohr's idea of the hydrogen atom violated Heisenberg's uncertainty principle because Bohr had precise orbitals, which are not allowed by Heisenberg. Schroedinger's wave equations gave distributions of probabilities rather than fixed locations. Even though they both predicted the same distance for the electron from the nucleus, they got there in fundamentally different ways.
Problems
31. Since the charge of a proton has the same magnitude as that of an electron, we can call them both e = +- 1.602x10-19 C (see pp. 278, 279 in the text). That makes the mass-to charge ratio of the proton mp/e = 1.044x10-8 kg/C. Then mp = (1.044x10-8 kg/C)(1.602x10-19 C) = 1.672x10-27 kg.
37. Recall that a weighted average is just the sum of the (value times the weight) for each of the numbers, all divided by the sum of the weights. Another way to say it is the summed products of values and their weights divided by the summed weights. For the mass spectrum of germanium, you must read the abundances from the graph, which will naturally introduce some uncertainty to the process because not everyone will read the heights of the bars in exactly the same way. Roughly speaking, we get [(69.9243 u)( 0.205) + (71.9217)(0.275) + (72.9234 u)(0.080) + (73.9219 u)(0.370) + (75.9214 u)(0.075) / (0.205 + 0.275 + 0.08 + 0.37 + 0.075) = 72.9920/1.005 = 72.6289 = 72.6 when rounded to three significant figures.
41. Since c = λν, λ = c/ν = (2.998x108 m s-1)/(992x103 Hz) = 302 m.
47. Use the formula E = hν
for both cases, but for the second case, where the wavelength rather than the
frequency is given, use the equation from problem 41 to transform the formula to
E = hc/λ.
First case: E = hν = (6.626x10-34 J
s)(7.72x1014 s-1) = 4.92x10-19 J
Second case: E = hc/λ = (6.626x10-34 J
s)(2.998x108 m s-1)/655x10-9 m) = 3.03x10-19
J
The photon of violet light (first case) is more energetic
than the photon of red light (second case).
53. To find the color, convert the frequency to wavelength
and compare with the band of wavelengths shown on page 287.
Since energy per photon is given rather than frequency, use E
= hc/λ from above and solve for λ:
λ = hc/E
λ = hc/E = [(6.626x10-34
J s)(2.998x108 m s-1)]/[(191 kJ/mol)/6.022x1023
mol-1)] = 6.27x10-7 m = 627 nm = orange.
55. Use the equation from page 296 for the difference
between two energy levels.
(a) ni = 5; nf = 1. ΔE
= B(1/ni2 - 1/nf2) = (2.179x10-18
J)(1/25 - 1/1) = -2.092x10-18 J → Ephoton = 2.092x10-18
J
(b) ni = 5; nf = 4. ΔE
= B(1/ni2 - 1/nf2) = (2.179x10-18
J)(1/25 - 1/16) = -4.90x10-20 J → Ephoton = 4.90x10-20
J
61. This problem has two steps. In the first, use same
formula as in problem 55, with ni = 1 and nf = ∞
to get the energy per photon. In the second step, multiply by the number of
photons per mole to get the energy per mole of photons, and adjust to kJ.
Step 1. ni = 1; nf = ∞.
ΔE = B(1/ni2 - 1/nf2)
= (2.179x10-18 J)(1/1 - 1/∞)
= -2.179x10-18 J → Ephoton = 2.179x10-18
J per photon
Step 2. Energy per mole of photons = Ephoton x
6.022x1023 photons mol-1 = (2.179x10-18
J per photon)(6.022x1023 photons mol-1) = 1.312x106
J mol-1 = 1.312x103 kJ mol-1.
63. Use De Broglie's equation λ
= h/mu, where h is Planck's constant, m is the mass of the particle (proton),
and u is the speed of the proton.
λ = h/mu = (6.626x10-34 J s)/[(1.67x10-27
kg)(2.55x106 m s-1)] = 1.56x10-13 m = 1.56x10-4
nm.
67. (a) The lowest principal shell for p orbitals is n = 2,
because l must be at least 1.
(b) For the f subshell, l must equal 3, which means n must be
at least 4.
71. Which of these combinations is permissible?
(a) n = 2, l = 1, ml = +1 Possible.
(b) n = 3, l = 3, ml = -3 Impossible because
l
cannot equal n.
(c) n = 3, l = 2, ml = -2 Possible.
(d) n = 0, l = 0, ml = 0 Impossible because
n may
not equal 0.
75. (a) n = ?, l = 2, ml = 0,
ms = ?
Since l = 2, n must be 3 or greater.
ms can always
be ±½.
Allowed orbitals are 3d, 4d,
5d, etc.
(b) n = 2, l = ?, ml = -1,
ms = -½.
l is bounded in two ways: because
n = 2, it must be 0 or 1,
and because ml = -1, it must be at least 1. Therefore, l = 1.
Allowed orbitals are 2p.
(c) n = 4, l = ?, ml = 2,
ms = ?
l is again bounded in two ways: up to 3 because
n = 4 and at least
2 because ml = 2. So l can be 2 or 3. ms can always be
±½.
Allowed orbitals are 4f and 4d.
(d) n = ?, l = 0, ml = ?,
ms = ½.
n can be 1 or higher because l = 0.
ml must be 0 because
l = 0.
Allowed orbitals are 1s, 2s,
3s, etc.
77. The weighted average for mercury is [(196x0.146) + (198x10.02) + (199x16.84) + (200x23.13) + (201x13.22) + (202x29.80) + (204x6.85)] / (0.146 + 10.032 + 16.86 + 23.13 + 13.22 + 29.80 + 6.95) = 200.7
83. The difference in energy between levels 1 and 2 is ΔE
= B(1/ni2 - 1/nf2) = B(1/1 - 1/4) =
(3/4)B.
The difference in energy between levels 1 and 3 is ΔE
= B(1/ni2 - 1/nf2) = B(1/1 - 1/9) =
(8/9)B.
From the equation of the form ΔE1-3 = CΔE1-2
(needed to compare with the four possible answers given in this problem), we get
C = ΔE1-3 /ΔE1-2 = (8/9)B/(3/4)B = 32/27.
This gives ΔE1-3 = (32/27)ΔE1-2, which
corresponds to answer (c).
89. The generalized equation will be ΔE = Z2B(1/ni2 - 1/nf2). To take a ground-state electron of helium completely off the atom (to ionize the atom), we will have Z = 2, ni = 1, and nf = ∞. Thus, ΔE = 4(2.179x10-18 J)(1/1 - 1/∞) = 8.716x10-18 J.
95. The minimum number of photons intercepted is the
detection limit for energy (in joules per second) divided by the energy per
photon (in joules per photon). The result will be the number of photons per
second.
Watts = joules per second = 4x10-21 J s-1
Energy per photon = hν
= (6.626x10-34 J s)(8.4x109 s-1) = 5.566x10-24
J photon-1
Photons per second = (4x10-21 J s-1)
/ (5.566x10-24 J photon-1)
= 7x102 photon s-1