2
2
" 1.388 ! 0.694 # " 0.694 #
R=$
% =$
% = 0.111 And T = 1 ! R = 0.889
& 1.388 + 0.694 ' & 2.082 '
(b) For electrons:
1/ 2
! 0.511 "
k1 = 1.388 #
$
% 938 &
1/ 2
= 0.0324
! 0.511 "
k2 = 0.694 #
$
% 938 &
= 0.0162
2
" 0.0324 ! 0.0162 #
R=$
% = 0.111 And T = 1 ! R = 0.889
& 0.0324 + 0.0162 '
No, the mass of the particle is not a factor. (We might have noticed that
could
be canceled from each term.)
7-1.
En1 n2 n3 =
E311 =
h 2! 2 2
n1 + n22 + n32
2mL2
(
)
(Equation 7-4)
h 2! 2 2 2 2
h 2! 2
3
+
1
+
1
=
11
E
where
E
=
0
0
2mL2
2mL2
(
)
(
)
(
)
E222 = E0 22 + 22 + 22 = 12 E0 and E321 = E0 32 + 22 + 12 = 14 E0
The 1st, 2nd, 3rd, and 5th excited states are degenerate.
E
(×E0)
14
321
12
222
311
10
221
8
6
211
4
111
2
s = 3/ 4
m
7-2.
En1 n2 n3 =
h 2! 2 " n12 n22 n32 # h 2! 2 " 2 n22 n32 #
+ %=
+ %
$ +
$ n1 +
2m & L12 L22 L23 ' 2mL12 &
4
9 '
n1 = n2 = n3 = 1 is the lowest energy level.
E111 = E0 (1 + 1 / 4 + 1 / 9 ) = 1.361E0 where E0 =
The next nine levels are, increasing order,
h 2! 2
2mL12
(Equation 7-5)
(Problem 7-2 continued)
7-3.
n1
n2
n3
E (! E0 )
1
1
2
1.694
1
2
1
2.111
1
1
3
2.250
1
2
2
2.444
1
2
3
3.000
1
1
4
3.028
1
3
1
3.360
1
3
2
3.472
1
2
4
3.778
(a) " n1 n2 n3 (x, y, z ) = A cos
n!z
n1! x
n!y
sin 2 sin 3
L
L
L
(b) They are identical. The location of the coordinate origin does not affect the
energy
level structure.
7-4.
" 111 (x, y, z ) = A sin
!x
!y
!z
sin
sin
L1
2 L1
3L1
" 112 (x, y, z ) = A sin
!x
!y
2! z
sin
sin
L1
2 L1
3L1
!x
!y
!z
sin
sin
L1
L1
3L1
!x
!y
2! z
" 122 (x, y, z ) = A sin sin
sin
L1
L1
3L1
" 121 (x, y, z ) = A sin
" 113 (x, y, z ) = A sin
!x
!y
!z
sin
sin
L1
2 L1
L1
7-5.
En1 n2 n3 =
n32 # h 2! 2 " 2 n22 n32 #
n22
h 2! 2 " n12
$ 2+
%=
+
+ %
$ n1 +
2m $ L1 (2 L1 )2 (4 L1 )2 % 2mL12 &
4 16 '
&
'
(from Equation 7-5)
"
n2 n2 #
h 2! 2
E0 = $ n12 + 2 + 3 % where E0 =
4 16 '
2mL12
&
(Problem 7-5 continued)
(a)
n1
n2
n3
E (! E0 )
1
1
1
1.313
1
1
2
1.500
1
1
3
1.813
1
2
1
2.063
1
1
4
2.250
1
2
2
2.250
1
2
3
2.563
1
1
5
2.813
1
2
4
3.000
1
1
6
3.500
(b) 1,1,4 and 1,2,2
7-7.
(
)(
2
)
)(
1.055 # 10"34 J s ! 2
h 2! 2
E0 =
=
= 37.68eV
2mL2 2 9.11 # 10"31 kg 0.10 # 10"9 m 2 1.609 # 10"19 J / eV
(
E311 ! E111 = "E = 11E0 ! 3E0 = 8 E0 = 301eV
(Problem 7-7 continued)
E222 ! E111 = "E = 12 E0 ! 3E0 = 9 E0 = 339eV
)
E321 ! E111 = "E = 14 E0 ! 3E0 = 11E0 = 415eV
7-8.
(a) Adapting Equation 7-3 to two dimensions (i.e., setting k3 = 0), we have
" n1 n2 = A sin
n1! x
n!y
sin 2
L
L
(b) From Equation 7-5, En1 n2 =
h 2! 2 2
n1 + n22
2mL2
(
)
(c) The lowest energy degenerate states have quantum numbers n1 = 1, n2 = 2,
and n1 = 2,
n2 = 1.
7-9.
(a) For n = 3, l = 0, 1, 2
(b) For l = 0, m = 0 . For l = 1, m = !1, 0, + 1 . For l = 2, m = !2, ! 1, 0, + 1, + 2 .
(c) There are nine different m-states, each with two spin states, for a total of 18
states for
n = 3.
7-10. (a) For l = 4
L = l (l + 1) h = 4 (5 ) h = 20 h
ml = 4h
!min = cos"1
4
# !min = 26.6°
20
(b) For l = 2
L = 6h
!min = cos"1
ml = 2h
2
# !min = 35.3°
6
7-12. (a)
+1
l =1
L = 2h
0
−1
(b)
+2
+1
0
l=2
L = 6h
(c)
+4
+3
+2
l=4
+1
L = 20h
0
−1
−2
−3
−4
(d) L = l (l + 1) h (See diagrams above.)
7-13.
2
(
)
L2 = L2x + L2y + L2z ! L2x + L2y = L2 " L2z = l (l + 1)h 2 " (mh ) = 6 " m 2 h 2
(a)
(L
+ L2y
)
= 6 ! 2 2 h 2 = 2h 2
(b)
(L
+ L2y
)
= 6 ! 0 2 h 2 = 6h 2
2
x
2
x
min
max
(
)
(
)
(c) L2x + L2y = (6 ! 1)h 2 = 5h 2
Lx and Ly cannot be determined separately.
(d) n = 3
7-15.
L= r? p
dL dr
dp
=
! p+ r!
dt dt
dt
dr
dp
! p = v ! mv = mv ! v = 0 and r !
= r ! F . Since for V =V(r), i.e., central
dt
dt
forces,
F is parallel to r, then r ! F = 0 and
dL
=0
dt
7-16. (a) For l = 3, n = 4, 5, 6, … and m = −3, −2, −1, 0, 1, 2, 3
(b) For l = 4, n = 5, 6, 7, … and m = −4, −3, −2, −1, 0, 1, 2, ,3 ,4
(c) For l = 0, n = 1 and m = 0
(d) The energy depends only on n. The minimum in each case is:
E4 = !13.6eV / n 2 = !13.6eV / 42 = !0.85eV
E5 = !13.6eV / 52 = !0.54eV
E1 = !13.6eV
7-17. (a) 6 f state: n = 6, l = 3
(b) E6 = !13.6eV / n 2 = !13.6eV / 62 = !0.38eV
(c) L = l (l + 1) h = 3 (3 + 1) h = 12 h = 3.65 " 10!34 J s
(d) Lz = mh
Lz = !3h, ! 2h, ! 1h, 0, 1h, 2h, 3h
(
)
7-20. (a) For the ground state, P (r )$r = ! 2 4" r 2 $r =
For "r = 0.03a0 , at r = a0 we have P (r )"r =
4r 2 #2 r / a0
e
$r
a03
4a02 !2
e (0.03a0 ) = 0.0162
a03
(b) For
2
"r = 0.03a0 , at r = 2a0 we have P (r )"r =
4 (2a0 )
3
0
a
e !4 (0.03a0 ) = 0.0088
7-21.
P (r ) = Cr 2 e !2 Zr / a0 For P(r) to be a maximum,
" $ 2 Z % !2 Zr / a0
#
dP
2 Zr $ a0
%
= C (r 2 * !
+ 2re !2 Zr / a0 ) = 0 & C '
! r + e !2 Zr / a0 = 0
+e
*
dt
a0 , Z
.( , a0 /)
This condition is satisfied with r = 0 or r = a0 /Z. For r = 0, P(r) = 0 so the
maximum
P(r) occurs for r = a0 /Z.
& ! 2!
7-22.
2
2 2
'" d# = ' ' ' " r sin$ drd$ d% = 1
0 0 0
#
2
#
2 2
= 4! +" r dr = 4! C
0
2
210
% Zr & 2 $ Zr / a
+0 ') a0 (* r e 0 dr = 1
"
$ Z 2 r 4 % # Zr / a0
2
= 4! C210
*0 &( a02 ') e dr = 1
Letting x = Zr / a0 , we have that r = a0 x / Z and dr = a0 dx / Z and
substituting
these above,
2
&" d# =
2 $
4! a03C210
4 %x
&0 x e dx
Z3
Integrating on the right side
!
4 "x
#x e
dx = 6
0
1/ 2
Solving for C
2
210
yields:
C
2
210
" Z3 #
Z3
=
$ C210 = %
3 &
24! a03
' 24! a0 (
7-26. For the most likely value of r, P(r) is a maximum, which requires that (see
Problem 7-24)
# % Z&
$
dP
= A cos2 ! ' r 4 ) " * e " Zr / a0 + 4r 3e " Zr / a0 ( = 0
dr
'- + a0 ,
(.
(
)
For hydrogen Z = 1 and A cos2 ! r 3 / a0 (4a0 " r )e " r / a0 = 0 . This is satisfied for
r=0
and r = 4a0. For r = 0, P(r) = 0 so the maximum P(r) occurs for r = 4a0.
7-33. (a) There should be four lines corresponding to the four mJ values −3/2, −1/2,
+1/2, +3/2.
(b) There should be three lines corresponding to the three ml values −1, 0, +1.
7-68.
P (r ) =
4 Z 3 2 !2 Zr / a0
r e
a03
(See Problem 7-63)
For hydrogen, Z = 1 and at the edge of the proton r = R0 = 10!15 m. At that point,
the
exponential factor in P(r) has decreased to:
e !2 R0 / a0 = e
(
!2 10!15
) (0.529"10
!10
m
) = e !(3.78"10 ) # 1 ! 3.78 " 10!5 # 1
!5
Thus, the probability of the electron in the hydrogen ground state being inside the
nucleus,
to better than four figures, is:
4r 2
P (r ) = 3
a0
r0
P = ! P (r )dr =
0
R0
!
0
4r 2
4
= 3
3
a0
a0
(
R0
4 r3
r
dr
=
!0
a03 3
R0
2
0
3
)
4 10!15 m
4 " R03 #
= 3% &=
a0 ' 3 ( 3 0.529 $ 10!10 m
(
3
)
= 9.0 $ 10!15
7-70. (a) Substituting ! (r," ) into Equation 7-9 and carrying out the indicated
operations
yields (eventually):
%
h2
h2
! (r," )#& 2 / r 2 % 1 / 4a02 $' % ! (r," ) %2 / r 2 + V! (r," ) = E! (r," )
2µ
2µ
(
)
Canceling ! (r," ) and recalling that r 2 = 4a02 (because ! given is for n = 2)
we
h2
!1 / 4a02 + v = E
have !
2µ
(
)
The circumference of the n = 2 orbit is:
C = 2! (4a0 ) = 2" # a0 = " / 4! = 1 / 2k.
Thus, #
h2 !
1 "
h2k 2
#
+
V
=
E
$
+V = E
2 µ %' 4 / 4k 2 &(
2µ
p2
+ v = E and Equation 7-9 is satisfied.
(b) or
2m
2
$
& r ' % r / a0
2
2
!
dx
=
A
,0
, (* a0 )+ e cos " r sin" drd" d# = 1
2
2
2
2!
& r ' % r / a0 2 !
2
A ,( ) e
r dr , cos " sin" d" , d# = 1
a
0* 0 +
0
0
$
2
Integrating (see Problem 7-22),
( )
A2 6a03 (2 / 3)(2! ) = 1
A2 = 1 / 8a03! " A = 1 / 8a03!