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Transcript
```Homework Solutions 2B
1st edition in bold, (2nd edition with parenthesis if different)
Q: 12,20,23,25,26 ; P: 1,5,10,14,19,21,27,43(47),68(70)
Q29.26. Reason: The amount of energy carried off by the photon may be determined by
E = hc/λ = (6.63 ×10
−34
J ⋅ s)(3.0 ×10 m/s)(eV/1.6 ×10
8
−19
−7
J)/ 2.75 ×10 m = 4.52 eV
The correct choice is E.
Assess:
The atom’s energy will decrease by the amount of energy carried off by the photon.
Q29.12. Reason: Case (a) represents a possible electron configuration for an atom with seven electrons (nitrogen) in the
ground state and the configuration is 1s22s22p3.
Case (b) does nor represent a possible electron configuration since it has three electrons in the 2s state and only two are
allowed.
Case (c) represents a possible electron configuration for an atom with five electrons (Boron) in an excited state and the
configuration is 1s2 2p3.
Assess:
The lower level states fill according to 1s22s22p63s23p6. . . .
Q29.20. Reason: Since the wavelengths are labeled with subscripts 1–7, let’s agree to refer to these as transition 1 through
7. Transition 1 is associated with λ1 , etc.
Transition 1 is not allowed because it violates the selection rule ∆l = ±1.
Transition 2 is in the emission spectrum and it is just out of the visible range at the blue end and into the UV.
Transition 3 is not allowed because it violates the selection rule ∆l = ±1.
Transition 4 is not allowed because it violates the selection rule ∆l = ±1.
Transition 5 is in the IR range of the emission spectrum.
Transition 6 is in the IR range of the emission spectrum.
Transition 7 is in the IR range of the emission spectrum.
Assess:
An energy level diagram provides a tremendous amount of information.
Q29.23. Reason: The electron gives up some of its energy to the atom. At atom in its ground state cannot emit a photon,
so the atom is first boosted to an excited state (one of the orbital electrons jumps to a higher state) and then it can emit a photon
as it drops to a lower state. If the excited electron drops all the way back to its lowest state (leaving the atom in the ground
state) then the energy of the photon is equal to the change in energy of the colliding electron. But the excited electron does not
have to drop all the way back down to its lowest state if it jumped more than one energy state; if it drops down to an
intermediate energy state then the energy of the emitted photon is less than the energy of the colliding electron that boosted up
the orbital electron. In this case ∆Eelec > Ephoton. Both cases are possible, so the correct choice is B.
Assess: Drawing a little energy-level diagram as you think this question through helps one keep
track of the various energies.
Q29.25. Reason: In the ground-state configuration each energy state will fill before the next one in energy fills. This
eliminates choices B and D. Choice A is eliminated because a p state can’t have 8 electrons. That leaves C.
The correct choice is C.
Assess:
We also count to verify that our choice C has 20 electrons.
P29.1. Prepare: Please refer to Figure 29.2b and Table 29.1, which gives the wavelengths of visible lines in the hydrogen
spectrum. We will use Equation 29.2, which is the Balmer formula.
Solve: (a) The wavelengths in the hydrogen emission spectrum are 656.6 nm, 486.3 nm, 434.2 nm, and 410.3 nm. The
formula for the Balmer series can be written
1 91.18 nm
1
− 2 =
2
λ
n
m
where m = 1, 2, 3, . . . and n = m + 1, m + 2, . . . For the first wavelength,
1
n2 − m 2
1 91.18 nm
1
= 0.1389
=
0.1389
⇒
=
−
n2m 2
m 2 n2 656.5 nm
This equation is satisfied when m = 2 and n = 3. For the second wavelength (486.3 nm) the equation is satisfied for m = 2 and n
= 4. Likewise, for the next two wavelengths m = 2 and n = 5 and 6.
(b) The fifth line in the spectrum will correspond to m = 2 and n = 7. Its wavelength is
λ=
Assess:
91.18 nm
( ) −( )
1 2
2
1 2
7
⎛ 196⎞
=⎜
⎟ (91.18 nm) = 397.1 nm
⎝ 45 ⎠
The Balmer wavelengths are in the visible region.
P29.5. Prepare: For a neutral atom, the number of electrons is the same as the number of protons. The atomic number of
Be, C, and N is 4, 6, and 7, respectively.
Solve: (a) Since Z = 4, the 9Be+ ion has 3 electrons, 4 protons, and 9 – 4 = 5 neutrons.
(b) Since Z = 6, 12C has 6 electrons, 6 protons, and 12 – 6 = 6 neutrons.
(c) Since Z = 7, the triply charged 15N+++ ion has 4 electrons, 7 protons, and 15 – 7 = 8 neutrons.
Assess:
We need a good understanding of the atomic model to label atoms and ions.
P29.10. Prepare: To conserve energy, the emission and the absorption spectra must have exactly the energy lost or gained
by the atom in the appropriate quantum jumps.
Solve: (a)
(b) The energy of a light quantum is E = hf = hc/λ. We can use this equation to find the emission and absorption wavelengths.
The emission energies from the above energy-level diagram are E2→1 = 4.0 eV, E3→1 = 6.0 eV,
and E3→2 = 2.0 eV. The wavelength corresponding to the 2→1 transition is
λ 2→1 =
hc (4.14 × 10−15 eV s)(3.0 × 108 m/s)
=
= 310 nm
E21
4.0 eV
Likewise, λ 3→1 = hc/E3→1 = 210 nm, and λ 2→1 = 620 nm.
(c) Absorption transitions start from the n = 1 ground state. The energies in the atom’s absorption spectrum are E1→2 = 4.0 eV and
E1→3 = 6.0 eV. The corresponding wavelengths are λ1→2 = hc/E1→2 = 310 nm and λ1→3 = hc/E1 →3 = 210 nm.
Assess: Only those transitions whose energies match the energy intervals between the stationary
states are observed as absorption or emission spectra.
P29.14. Prepare: The kinetic energy of an electron in the n = 1 state may be determined by KE = mv12/ 2, where
6
v1 = 2.19 ×10 m/s. The
electric
potential
energy
of
the
atom
in
the n = 1 state
may
−11
be
determined
by PE = ke /r1, where r1 = 5.2 9 × 10 m.
Solve: The ration of the kinetic energy to the electric potential energy for a hydrogen atom in the n = 1 state may be
determined by
2
KE mv 21 /2 mv21r1 (9.11 × 10−31 kg)(2.19 ×106 m/s)2 (5.29 ×10−11 m)
= 0.50
=
=
=
2(8.99 × 109 N⋅ m2 /C 2 )(1.60 ×10−19 C) 2
2ke 2
PE ke 2/r1
2
Assess: The kinetic energy is essentially half the electric potential energy when a hydrogen atom is
in the n = 1 state.
P29.19. Prepare: We will use Equation 29.2 or 29.18 with λ0 = 91.18 nm.
Solve: Photons emitted from the n = 4 state start in energy level n = 4 and undergo a quantum jump to a lower energy level
with m < 4. The possibilities are 4 → 1, 4 → 2, and 4 → 3. According to Equation 29.18, the transition 4 → m emits a photon
of wavelength.
λ=
λ0
(1/m2 − 1/n2 )
=
91.18 nm
(1/m2 −1/16)
These values are given in the table below.
Transition
Wavelength
4→1
97.3 nm
486 nm
4→2
1876 nm
4→3
Assess: Transitions are in the ultraviolet, visible, and infrared regions.
P29.21. Prepare: No two electrons can have exactly the same set of quantum numbers (n, l, m, ms).
For n = 1, there are a total of 2 states with the quantum numbers given by (1, 0, 0, ± 12).
Solve:
For n = 2, there are a total of 8 states:
⎛
1⎞
⎜ 2, 0 , 0, ± ⎟
⎝
2⎠
⎛
1⎞
⎜2, 1, − 1, ± ⎟
⎝
2⎠
⎛
1⎞
⎜ 2,1, 0, ± ⎟
⎝
2⎠
⎛
1⎞
⎜2, 1, 1, ± ⎟
⎝
2⎠
For n = 3, there are a total of 18 states:
1⎞
⎛
⎜ 3, 0, 0, ± ⎟
2⎠
⎝
1⎞
⎛
⎜ 3, 1, − 1, ± ⎟
2⎠
⎝
⎛
1⎞
⎜ 3, 2, 2, ± 2 ⎟
⎝
⎠
⎛
1⎞
⎜ 3, 2, 1, ± 2 ⎟
⎝
⎠
1⎞
⎛
⎜ 3, 1, 0, ± ⎟
2⎠
⎝
⎛
1⎞
⎜ 3, 2, 0, ± 2 ⎟
⎝
⎠
1⎞
⎛
⎜ 3, 1, 1, ± ⎟
2⎠
⎝
⎛
1⎞
⎜ 3, 2, − 1, ± 2 ⎟
⎝
⎠
⎛
1⎞
⎜ 3, 2, − 2, ± 2 ⎟
⎝
⎠
Assess: Learn well the properties of the four quantum numbers.
P29.27. Solve: Si, Ge, and Pb are all in the same column of the periodic table as carbon. The electron confi-guration of Si
(Z = 14) is 1s22s22p63s23p2. The electron configuration of Ge (Z = 32) is
1s22s22p63s23p64s23d104p2
The electron configuration of Pb (Z = 82) is
1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2
Assess: It is necessary to recall that the 4f subshell fills before the 5d subshell and after 6s. In Ge,
the 4s subshell fills before the 3d subshell.
P29.43. Prepare: For a neutral atom, the number of electrons is the same as the number of protons, which is the atomic
number Z. An atom’s mass number is A = Z + N, where N is the number of neutrons. Mass of the proton is 1.67 ×1 0
its volume is 4π / 3(7.0 ×10
−15
−27
kg and
3
m) .
Au atom, Z = 79. So, N = 197 – 79 = 118. A neutral 197Au atom contains 79 protons, 79 electrons, and 118
197
Solve: (a) For a
neutrons.
(b) Assuming that the neutron rest mass is the same as the proton rest mass, the density of the gold nucleus is
ρ nucleus =
197 mproton
197(1.67 ×10 −27 kg)
17
3
=
−15
3 = 2. 3 × 10 kg/m
4π
−15
3
4π
(7.0
×10
m)
(7.0
×10
m)
3
3
(c) The nuclear density in part (b) is 2.0 × 1013 times the density of lead.
Assess: The mass of the matter is primarily in the nuclei and the volume of the matter is essentially
due to the electrons around the nuclei.
P29.68. Prepare: Please refer to Figure P29.68. We need to use the condition ∆l = 1 to determine the allowed transitions,
and the equation
∆E =
hc
3
λ
=
1240 eV nm
λ
to find the wavelengths.
Solve: (a)
Transition
6p → 6s
6d → 6p
7s → 6p
7p → 6s
7p → 7s
8s → 6p
8s → 7p
8p → 6s
8p → 7s
8p → 8s
8p → 6d
∆E (eV)
λ (nm)
6.70
2.14
1.22
8.84
0.92
2.52
0.38
9.53
1.61
0.31
0.69
185
579
1016
140
1350
492
3260
130
770
4000
1800
(b) From part (a), the 492-nm-wavelength blue emission line in the Hg spectrum corresponds to the transition
8s → 6p. The energy of the 8s state is 9.22 eV. The atom is excited to the 8s state from the ground state. Thus the minimum
kinetic energy of the electron must be
1
2
mv2 = 9.22 eV ⇒ v =
2(9.2 2 eV)(1.60 × 10 −19 J/eV)
9.11 × 10 −3 1 kg
4
= 1.8 × 10 m /s
6
```
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