Key 9
1. Key to Prob. 7.1
! 
E=
h̄2 π 2  n1
2m
Lx
2
n2
+
Ly
!2
n3
+
Lz
2

.
Now Lx = L, Ly = Lz = 2L and let E0 = h̄2 π 2 /8mL2 . Then ,
E = E0 4n21 + n22 + n23 .
Ground state corresponds to n1 = n2 = n3 = 1. Then E = 6E0 ..
There are two degenerate first excited states – one for n1 = 1, n2 = 2, n3 = 1
and the other one for n1 = 1, n2 = 2, n3 = 2. The energy is E = 9E0 .
The next excited state correspond to n1 = 1, n2 = 2, n3 = 2 with energy
E = 12E0 .
There are next two degenerate excited states – one for n1 = 1, n2 = 1, n3 = 3
and the other one for n1 = 1, n2 = 3, n3 = 1. with energy E = 14E0 .
The next excited state correspond to n1 = 2, n2 = 1, n3 = 1 with energy
E = 18E0 .
There are next two degenerate excited states – one for n1 = 2, n2 = 1, n3 = 2
and the other one for n1 = 2, n2 = 2, n3 = 1. with energy E = 21E0 .
The next excited state correspond to n1 = 2, n2 = 2, n3 = 2 with energy
E = 24E0 .
Therefore the first six lowest states are ψ111 , ψ121 , ψ112 , ψ122 , ψ113 , and ψ131 with
relative energies E/E0 = 6, 9, 9, 12, 14, 14. First and third excited states are
doubly degenerate.
1
2. Key to Prob. 7.5
(a) n1 = n2 = n3 = 1 and
E111 =
3h2
3(6.63 × 10−34 )2
=
= 2.47 × 10−13 J ≈ 1.54 MeV.
2
−27
−28
8mL
8 × 1.67 × 10
× 4 × 10
(b) States 211, 121, 112 have the same energy and
E=
(22 + 12 + 12 )h2
8mL2
!
= 2E111 ≈ 3.08 MeV,
and states 221, 122, 212 have the energy
E=
(22 + 22 + 12 )h2
8mL2
!
= 3E111 ≈ 4.63 MeV.
(c) Both states are threefold degenerate.
3. Key to Prob. 7.12
(a) The wavefunction exponetially decays to zero.
(b) The probability of finding the electron in a volume element dV is given
by |ψ|2 dV . Since the wavefunction has spherical symmetry (does not depend
on the angles θ and φ), the volume element can be identified with the volume
of a spherical shell of radius r, dV = 4πr2 dr. The probability of finding the
electron between r and r + dr (i.e. within the spherical shell) is P = |ψ|2 dV =
4πr2 |ψ|2 dr.
(c) Look in your text book.
(d)
Z
|ψ|2 dV
= 4π
Z
|ψ|2 r2 dr,
!
1 Z ∞ −2r/a0 2
1
e
r dr,
= 4π
π
a30 0
!
4 Z ∞ −2r/a0 2
=
e
r dr,
a30 0
!
!
4
a30
=
= 1.
a30
4
2
(e) P = 4π
R r2
r1
|ψ|2 r2 dr where r1 = a0 /2 and r2 = 3a0 /2.
4 Z r2 2 −2r/a0
P = 3
r e
dr,
a0 r 1
1 Z 3 2 −z
2r
=
z e dz, (where z = ).
2 1
a0
1 2
= − z + 2z + 2 e−z |31 = 0.496.
2
4. Key to Prob. 7.22
1
ψ2s (r) = √
4 2π
1
a0
3/2 2−
r
e−r/2a0 .
a0
At r = a0 = 0.529 × 10−10 m, we find ψ2s (a0 ) = 9.88 × 1014 m−3/2 .
(b) |ψ2s (a0 )|2 = 9.75 × 1029 m−3 .
(c) Using the result to part (b), we get
P2s (a0 ) = 4πa20 |ψ2s (a0 )|2 = 3.43 × 1010 m−1 .
5. Key to Prob. 7.23
√ 5/2
R2p (r) = Are−r/2a0 where A = 1/{2 6a0 }.
2
P (r)r2 R2p
(r) = A2 r4 e−r/a0 .
So,
hri =
Z
0
∞
rP (r)dr = A
2
Z
0
∞
r5 e−r/a0 = A2 a60 5! = 5a0 = 2.645Å.
3