8-24
P1s r 
4 2 2r
r e
a30
a0
for hydrogen ground state, U r  

U   U rP1 srdr  
0

4ke a0 
 
a03  2 
 re 2r a 0 dr
0
2r
z
 ze dz where z  a
0
0
ke
 213.6 eV   27.2 eV.
a0
2
To find K , we note that K  U  E  
8-25
2 
2
2

4ke
3
a0
ke2
is potential energy ( Z  1 )
r
ke2
ke2
 13.6 eV so, K 
 13.6 eV .
2 a0
a0
The most probable distance is the value of r which maximizes the radial probability density
Pr  rRr . Since Pr is largest where rRr  reaches its maximum, we look for the most
2
probable distance by setting
drRr
dr
equal to zero, using the functions Rr from Table 8.4.
For clarity, we measure distances in bohrs, so that
r
becomes simply r, etc. Then for the 2s
a0
state of hydrogen, the condition for a maximum is
0
2r  r e  2  2r  12 2r  r e
d
dr
2
r 2
2
r 2
or 0  4  6r  r 2 . There are two solutions, which may be found by completing the square to
get 0  r  3   5 or r  3  5 bohrs . Of these r  3  5  5.236a0 gives the largest value of
Pr , and so is the most probable distance. For the 2p state of hydrogen, a similar analysis

1 2  r 2
d 2 r 2
r e
 2r  r e
with the obvious roots r  0 (a minimum) and r  4 (a
gives 0 
2 
dr

maximum). Thus, the most probable distance for the 2p state is r  4a0 , in agreement with the
simple Bohr model.
2

8-26

The probabilities are found by integrating the radial probability density for each state, P(r),
from r  0 to r  4a0 . For the 2s state we find from Table 8.4 (with Z  1 for hydrogen)
2
Changing variables from r to z 
integration by parts gives

2
2
2
4a 0
 r  
r  
r  r a
r  r a
1
  2   e 0 and P  8a0     2   e 0 dr .
a0   a0 
a0 
0  a0  
1 
P2 sr  rR2 s r  8a0 
2


4
r
gives P  8 1  4z 2  4z 3  z 4 e z dz . Repeated
a0
0

 


P  8 1  4 z 2  4 z3  z 4  8z 12 z 2  4 z3  8  24 z  12z2  24  24z   24  e z
8
1
64  96  104  72  24 e
4

 8  0.176
4
0
For the 2p state of hydrogen P2 p r  rR2p r  24 a0 
P  24a0 
1
4a 0
4
 r 
4
1 
2
r  r a
  e 0 and
a0 
4
r a
1
4 z
 e 0 dr  24  z e dz . Again integrating by parts, we get
 


a
0
0
0


4
P  24 1 z 4  4z 3 12z 2  24z  24 e z
0


 24 1 824e 4  24  0.371 . The probability for the
2s electron is much smaller, suggesting that this electron spends more of its time in the outer
regions of the atom. This is in accord with classical physics, where the electron in a lower
angular momentum state is described by orbits more elliptic in shape.
8-29
To find r we first compute r2
hydrogen: P1 s r 
r2 
4
a30
2 2r a 0
r e

. Then r2   r 2 P1 s rdr 
0
r
4

2r
4 2 r a
 r e 0 dr . With z  a , this is
a30 0
0
5

4 a0 
4 z
4 z

 z e dz . The integral on the right is (see Example 8.9)  z e dz  4 ! so that
3 


2
a0
0
0

5
2
using the radial probability density for the 1s state of
4 a0 
2
2
 3   4!  3 a0 and r  r  r


2
a0
  3 a
2 12
2
0
 1.5 a0 
2

12
 0.866a0 . Since r is an
appreciable fraction of the average distance, the whereabouts of the electron are largely
unknown in this case.
8-30
The averages r and r2 are found by weighting the probability density for this state
 Z  2 2 Zr
P1 s r  4 3 r e
a0 
a0
with r and r , respectively, in the integral from r  0 to r   :
2

 Z  3 2 Zr
r   rP1 s rdr  4  3  r e
a0 0
0
a0
dr

 Z 
r2   r 2 P1 s rdr  4  3  r 4 e 2 rZ
a0 0
0
Substituting z 
a0
dr
2 Zr
gives
a0
3
 Z   a0 4  3  z
3!  a  3 a 
r  4      z e dz   0    0 
a
4  Z  2  Z 
2
Z


 0 
0
3
 Z   a 
r2  4    0 
a0  2 Z 


5
12
a0  9 
3  
Z 
 4 
from the average potential energy
and r  r
2
 r

U  kZe 2 
0
2 12

4 z
 z e dz 
0
2
2
4!  a0 
 a0 
   3 
8  Z 
 Z 
a0 
 0.866 . The momentum uncertainty is deduced
 Z 
3
 Z  
1
P1 srdr  4kZe 2    re 2 Zr
 a0  0
r
 Z   a 
k Ze 
 4kZe 2    0   
.
a0  2Z 
a0
3
a0
2
2
k Ze 
2
for the 1s level, and a0 
, we obtain
Then, since E  
2a0
m e ke2

2
2m e k Ze  Z  
 2me K  2me E  U  
   .
 a0 
2a0
2
2
p

2
 
With p  0 from symmetry, we get p  p

consistent with the uncertainty principle.
E  2BB  hf
2
9-1



12

Z
and rp
 0.866 for any Z,

a0

2 9.27  10 24 J T 0.35 T   6.63  10 34 Js f so f  9.79  10 9 Hz
9-4
(a)
3d subshell  l  2  m l  2, 1, 0, 1, 2 and m s  
l
ml
ms
2
2
2
2
2
2
2
2
2
2
–2
–2
–1
–1
0
0
1
1
2
2
–1/2
+1/2
–1/2
+1/2
–1/2
+1/2
–1/2
+1/2
–1/2
+1/2
1
for each m l
2
n
3
3
3
3
3
3
3
3
3
3
(b)
3p subshell: for a p state, l  1 . Thus m l can take on values l to l, or –1, 0, 1. For each
1
m l , m s can be  .
2
l
ml
ms
1
1
1
1
1
1
–1
–1
0
0
1
1
–1/2
+1/2
–1/2
+1/2
–1/2
+1/2
n
3
3
3
3
3
3
9-6
The exiting beams differ in the spin orientation of the outermost atomic electron. The energy
difference derives from the magnetic energy of this spin in the applied field B:
 e 
U   s  B  g S zB   gB Bm s .
2m 

1 
With g  2 for electrons, the energy difference between the up spin m s   and down spin

2 
1 

m    orientations is

 s
2 


   2
2
2
From Equation 8.9 we have E  
2 n1  n 2  n 3 
2mL 
U  gBB  2  9.273  10 24 J T 0.5 T   9.273  10 24 J  5.80  10 5 eV .
2
9-17
2

1.054  10 34   2 n12  n22  n32   1.5  10 18 J n 2  n 2  n 2  9.4 eV n 2  n 2  n 2
E

 1 2 3 
 1 2 3
2
2 9.11  10 31 2  10 10 
2
(a)
2 electrons per state. The lowest states have
n12  n22  n32  1, 1, 1  E111  9.4 eV12  12  12  eV  28.2 eV .


For n12  n22  n32  1, 1, 2 or 1, 2, 1 or (2, 1, 1),
2
E112  E121  E211  9.4 eV 1  1  2
2
2
 56.4 eV
E min  2  E111  E112  E121  E211   2 28.2  3  56.4  398.4 eV
(b)


All 8 particles go into the n12  n22  n32  1, 1, 1 state, so
E min  8  E111  225.6 eV .
9-21
2
2
4
(a)
1s 2s 2p
(b)
For the two 1s electrons, n  1 ,
l  0 , ml  0 , ms  
For the two 2s electrons, n  2 , l  0 , m l  0 , m s  
1
.
2
1
.
2
For the four 2p electrons, n  2 , l  1 , m l  1, 0 , 1 , m s  
1
.
2
!
9-15
The spin of the atomic electron has a magnetic energy in the field of the orbital moment
$ e '
given by Equations 9.6 and 9.12 with a g-factor of 2, or U = "µ s # B = 2 &
S B = 2µBmsB .
% 2me )( z
The magnetic field B originates with the orbiting electron. To estimate B, we adopt the
equivalent viewpoint of the atomic nucleus (proton) circling the electron, and borrow a
result from classical electromagnetism for the B field at the center of a circular current
2k µ
2
loop with radius r: B = m3 . Here km is the magnetic constant and µ = i" r is the
r
magnetic moment of the loop, assuming it carries a current i. In the atomic case, we
identify r with the orbit radius and the current i with the proton charge +e divided by the
2" r
evr " e %
orbital period T =
. Then µ =
=
L where L = me vr is the orbital angular
2 $# 2me '&
v
1 2
momentum of the electron. For a p electron l = 1 and L = [ l(l + 1) ] h = 2h , so
" eh %
µ =$
2 = µB 2 = 1.31 ( 10 )23 J T . For r we take a typical atomic dimension, say
# 2me '&
(
)
4 a0 = 2.12 " 10 #10 m for a 2p electron, and find
B=
Since ms is ±
(
)(
) = 0.276 T .
2 10 "7 N A 2 1.31 # 10 "23 J T
3
(2.12 # 10 "10 m )
1
the magnetic energy of the electron spin in this field is
2
(
)
U = ± µB B = ± 9.27 " 10 #24 J T (0.276 T ) = ±2.56 " 10 #24 J = ±1.59 " 10 #5 eV .
The up spin orientation (+) has the higher energy; the predicted energy difference
between the up (+) and down (–) spin orientations is twice this figure, or about
!
3.18 " 10 #5 eV —a result which compares favorably with the measured value,
5 " 10 #5 eV .
!
!
!
9-24
Ato
m
3s
3p
4s
Electr on
Co nfig urat ion
Na
[Ne]3s1
Mg
[Ne]3s2
Al
[Ne]3s23p1
Si
[Ne]3s 3p
P
[Ne]3s23p3
S
[Ne]3s23p4
Cl
[Ne]3s23p5
Ar
[Ne]3s23p6
K
[Ar]4s1
2
2
The 3s subshell is energetically lower and so fills before the 3p. According to Hund’s rule,
electrons prefer to align their spins so long as the exclusion principle can be satisfied.