Physics 43 Chapter 41 Homework #10 Key
1. Show that the penetration distance η has the units of m.
From Equation 40.41, the units of the penetration distance are

J×s
(kg × m 2 /s 2 ) × s
⇒
=
=
2m(U 0 − E )
kg × J
kg × kg m 2 /s 2
η=
kg × m 2 /s
kg 2 × m 2 /s 2
=
kg × m 2 /s
= m
kg × m/s
2. (Exercise 6 in Knight Ch 40)
40.6. Solve: (a)
(b) For n = 2, the probability of finding the particle at the center of the well is zero. This is because the wave function is
zero at that point.
(c) This is consistent with standing waves. The n = 2 standing wave on a string has a node at the center of the string.
3. What is the probability that an electron will tunnel through a 0.45nm gap from a metal to a STM probe
if the work function is 4.0eV?
The work function E0 = 4.0 eV is the energy barrier U0 – E that an electron must either go over (photoelectric effect) or
tunnel through. From Equation 41.40, the electron’s penetration distance is
1.05 ×10−34 J s
=
=
9.72 ×10−11 m =
0.0972 nm
2m (U 0 − E )
2 ( 9.11×10−31 kg )( 4.0 eV ×1.6 ×10−19 J/eV )
η=

The tunneling probability through the barrier is
nm )
wη
Ptunnel
= e −2=
e −2( 0.50 nm ) ( 0.0972=
3.40 ×10−5
4. An electron has a kinetic energy of 12.0 eV. The electron is incident upon a rectangular barrier of height
20.0 eV and thickness 1.00 nm. By what factor would the electron’s probability of tunneling through the
barrier increase assuming that the electron absorbs all the energy of a photon with wavelength 546 nm
(green light)?
The original tunneling probability is T = e−2CL where
E) )
( 2m(U − =
12
C
=

2π ( 2 × 9.11× 10−31 kg ( 20 − 12) 1.6 × 10−19 J)
= 1.448 1× 1010 m −1
6.626 × 10−34 J⋅ s
12
hc 1 240 eV ⋅ nm
=
= 2.27 eV , to make the electron’s new kinetic energy
546 nm
λ
12 + 2.27 =
14.27 eV and its decay coefficient inside the barrier
The photon energy is hf
=
2π ( 2 × 9.11× 10−31 kg ( 20 − 14.27 ) 1.6 × 10−19 J)
=
C′
= 1.225 5 × 1010 m −1
6.626 × 10−34 J⋅ s
12
Now the factor of increase in transmission probability is
−9
10
e−2C′L
− C′ )
m −1
4.45
e2×10 m ×0.223×10 =
e=
= e2L ( C=
85.9
−2CL
e
5. An electron having total energy E = 4.50 eV approaches a rectangular energy barrier with U = 5.00 eV
and L = 950 pm as. Classically, the electron cannot pass through the barrier because E < U. However,
quantum-mechanically the probability of tunneling is not zero. Calculate this probability, which is the
transmission coefficient.
=
C
2 ( 9.11× 10−31 ) ( 5.00 − 4.50) ( 1.60 × 10−19 ) kg ⋅ m s
= 3.62 × 109 m −1
1.055 × 10−34 J⋅ s
T =e−2CL =exp  −2 ( 3.62 × 109 m −1 )( 950 × 10−12 m )  =exp ( −6.88)
=
T
1.03 × 10−3
6. (Knight Ch 40 E. 20)
40.20. Solve: Electrons are bound inside metals by an amount of energy called the work function E0 . This is the energy
that must be supplied to lift an electron out of the metal. In our case, E0 is the amount of energy (U 0 − E ) appearing in
Equation 40.41 for the penetration distance. Thus,
=
η

=
2m(U 0 − E )
(1.05 × 10−34 J s)
2(9.11 × 10
−31
kg)(4.0 eV × 1.60 × 10
−19
= 9.72 × 10−11 m
J/eV)
The probability that an electron will tunnel through a w = 4.54 nm gap (from a metal to an STM probe) is
Ptunnel =e−2 w/η =e−2(0.45×10
−9
m)/(9.72 ×10−11 m)
=9.5 × 10−5 =0.0095%
7. A one-dimensional harmonic oscillator wave function is
ψ = Axe −bx
2
(a) Find b and the total energy E. (b) Is this a ground state or a first excited state?
dψ
ψ = Axe− bx so= Ae− bx − 2bx 2 Ae− bx
dx
2
2
2
and
2
2
2
d2ψ
=
−2bxAe− bx − 4bxAe− bx + 4b2 x3 e− bx =
−6bψ + 4b2 x 2ψ
2
dx
Substituting into the Schrödinger equation,
 2mE 
 mω  2
−6bψ + 4b2 x 2ψ =
−
ψ + 
 xψ
  
  
2
For this to be true as an identity, it must be true for all values
of x.
So we must have both
(a)
Therefore
(b)
and
(c)
The wave function is that of the
b=
E
=
−6b =
−
mω 
2mE
and 4b2 = 

2

  
mω
2
3b 2
=
m
3
ω
2
first excited state .
8. Find the expectation value for the for the first two states of a harmonic oscillator.
∞
(a)
=
x0
12
2
 a
=
x   e− ax dx
π 
−∞
∫
∞
∫
∞
x
01
since the integrand is an odd function of x.
12
 4a3 
2 − ax 2
x
=
 x e dx
 π 
−∞
(b)
x1
=
(c)
0 ,
=
∫
−∞
0 , since the integrand is an odd function of x.
1
1
1
2
x (ψ 0 + ψ 1 ) dx =
x + x +
2
2 0 2 1
∞
∫ xψ
0
( x )ψ 1 ( x ) dx
−∞
The first two terms are zero, from (a) and (b). Thus:
2
9. The ground-state wave function for the electron in a hydrogen atom is
ψ (r ) =
1
π a0
3
e − r / a0
where r is the radial coordinate of the electron and a0 is the Bohr radius. (a) Show that the wave function
as given is normalized. (b) Find the probability of locating the electron between r1 = a0/2 and r2 = 3a0/2.
(a)
∞
 1 ∞
2
2
=
π ∫ ψ r 2dr 4π  3  ∫ r 2e−2r
∫ ψ dV 4=
 π a0  0
0
a0
2 
Using integral tables, ∫ ψ dV =
− 2  e−2r
a0 
a0
2
dr
∞
 2
 2   a02 
a02  
+
+
=
1
r
a
r

0

− 2 − 2  =
2  

 a0  

0
so the wave function as given is normalized.
3a0 2
 1  3a0 2
2
(b) =
π ∫ ψ r 2dr 4π  3  ∫ r 2e−2r
Pa0 2→ 3a0 2 4=
 π a0  a0 2
a0 2
a0
dr
Again, using integral tables,
Pa0 2→ 3a0 2
2 
=
− 2  e−2r
a0 
3a 2
a0
0
 2
a02  
2  −3  17a02  −1  5a02  
+
+
=
−
r
a
r
0.497 .

e 
0

 −e  4  =
2  
a02   4 


 
a0 2
10.
List the possible sets of quantum numbers for electrons in (a) the 3d subshell and (b) the 3p
subshell.
(a)
In the 3d subshell, n = 3 and  = 2 ,
we have
n

m
ms
3
2
+2
+1/2
3
2
+2
–1/2
3
2
+1
+1/2
3
2
+1
–1/2
3
2
0
+1/2
3
2
0
–1/2
(A total of 10 states)
(b)
In the 3p subshell, n = 3 and  = 1 ,
we have
n
3
1

+1
m
+1/2
ms
(A total of 6 states)
3
1
+1
3
1
+0
3
1
+0
3
1
–1
3
1
–1
–1/2
+1/2
–1/2
+1/2
–1/2
3
2
–1
+1/2
3
2
–1
–1/2
3
2
–2
+1/2
3
2
–2
–1/2
11. Find all possible values of L, Lz, and θ for an electron in a 3d state of hydrogen.
For a 3d state,
n = 3 and  = 2
Therefore,
L=
 (  + 1) =
m can have the values
so
6 = 2.58 × 10−34 J⋅ s
–2, –1, 0, 1, and 2
Lz can have the values − 2 , −  , 0,  and 2 .
Using the relation cosθ =
Lz
L
we find the possible values of θ
145°, 114°, 90.0°, 65.9°, and 35.3°
12
(a) Write out the electronic configuration for the ground state of oxygen (Z = 8). (b) Write out a set
of possible values for the quantum numbers n, ℓ, m ℓ, and ms for each electron in oxygen.
(a)
(b)
1s2 2s2 2p4
For the 1s electrons,
n = 1 ,  = 0 , m = 0 ,
−
For the two 2s electrons,
For the four 2p electrons,
and
ms = +
1
2
and
ms = +
1
2
or
1
.
2
n = 2 ;  = 1 ; m = −1 , 0, or 1; and
−
1
2
1
.
2
n = 2 ,  = 0 , m = 0 ,
−
ms = +
1
.
2