Physics 43 HW 17 Ch 42
E: 10, 11
P: 27, 28, 31, 33, 35
42.10. Solve: (a) Nine electrons (Z = 9) make the element fluorine (F). These are the nine lowest energy
states, so this is the ground state of F.
(b) Thirty-one electrons (Z = 31) make the element gallium (Ga). States fill in the order 4s – 3d – 4p. So
3d104s24p, with filled 3d and 4s shells, has the 31 electrons in the lowest possible energy states. This is the
ground state of Ga.
42.11. Solve: (a) Ten electrons (Z = 10) make the element neon (Ne). These are not the ten lowest energy
states because 1s22s22p6 would be lower in energy than 1s22s22p53d. This is an excited state of Ne.
(b) Twenty-six electrons (Z = 26) make the element iron (Fe). These are not the 26 lowest energy states because
the 3d shell is not filled. This is an excited state of Fe.
42.27. Solve: A hydrogen atom in its fourth excited state is in the n = 5 state. The energy of the wavelength
of the emitted photon is
hc 1240 eV nm
ΔE =
=
= 0.967 eV
λ
1282 nm
From Equation 42.2, the hydrogen atom’s energy must be one of the following values
13.60 eV
En = −
n2
The transition 5 → n corresponds to the energy
⎛1 1⎞
ΔE = −13.60 eV ⎜ 2 − 2 ⎟ = 0.967 eV ⇒ n = 3
⎝5 n ⎠
This means the angular momentum quantum number is l = 2, 1, or 0. The atom’s maximum possible orbital
angular momentum after the emission is
L = l ( l + 1)h = 2 ( 2 + 1)h = 6h
42.28. Solve: (a) From Equation 42.7, the radial wave function of hydrogen in the 1s state is
R1s ( r ) =
1
πa
3
B
e− r aB ⇒ R1s ( 12 aB ) =
1
−1
πa
3
B
e 2=
0.342
aB3
(b) From Equation 42.10, the probability density is
2
⎛a ⎞
Pr ( r ) = 4π r 2 Rnl ( r ) ⇒ P1s ( 12 aB ) = 4π ⎜ B ⎟
⎝ 2 ⎠
2
2
0.368
⎛ 1
−1 ⎞
e 2⎟ =
⎜⎜
3
⎟
aB
a
π
B
⎝
⎠
42.31. Solve: From Equation 42.7, the 2p radial wave function of the hydrogen atom can be written
⎛ r ⎞ − r 2 aB
R2 p ( r ) = A2 p ⎜
⎟e
⎝ 2aB ⎠
The normalization condition for the three-dimensional hydrogen atom is
∫
∞
0
∞
∞
⎛ r2 ⎞ −
2
Pr ( r ) dr = 4π ∫ r 2 Rnl ( r ) dr = 4π ∫ A22p ⎜ 2 ⎟ e 2 aB r 2dr
0
0
⎝ 4aB ⎠
2r
=π
A22p
aB2
∫
∞
0
r 4e − r aB dr =
A22pπ ⎡ 4! ⎤
1
⎢
⎥ = 24π aB3 A22p = 1 ⇒ A2 p =
24π aB3
aB2 ⎢⎣ (1 aB )5 ⎥⎦
42.33. Solve: (a) From Equation 42.7, the 2p radial wave function is
R2 p ( r ) =
A2 p
2 aB
re− r 2 aB
The graph of R2p(r) is seen to have a single maximum.
(b) R2p(r) is a maximum at the point where dR2p/dr = 0. The derivative is
dR2 p
dr
=
A2 p ⎡ − r 2 aB
r − r 2 aB ⎤ A2 p ⎡
r ⎤ − r 2 aB
e
−
⎢e
⎥=
⎢1 −
⎥e
2aB ⎣
2aB
⎦ 2aB ⎣ 2aB ⎦
The derivative is zero at r = 2aB, so R2p(r) is a maximum at r = 2aB .
(c) The radial wave function R2p and the probability density P2p(r) is smaller at r = 4aB than it is at r = 2aB. We
are less likely to find the electron at a point with r = 4aB than at a point with r = 2aB. However, there are many
more
points
with
r = 4aB than with r = 2aB. The increased number of points more than compensates for the decreased probability
per point. As a result, it is more probable to find the electron at distance r = 4aB than it is at distance r = 2aB.
B
B
B
B
B
B
B
B
B
B
42.35. Solve: The electron configuration in the ground state of K (Z = 19) is
1s22s22p63s23p64s1
This means that all the states except the 4s state are completely filled. For Ti (Z = 22), all the states up to the 4s
state are completely filled. The electron configuration is
1s22s22p63s23p64s23d2
The d subshell has only two electrons. In the case of Fe (Z = 26), the 3d subshell has six electrons and all the
lower level states are completely filled. The electron configuration is
1s22s22p63s23p64s23d6
The ground-state electron configurations of Ge (Z = 32) and Br (Z = 35) are
1s22s22p63s23p64s23d104p2
1s22s22p63s23p64s23d104p5