Phy209-Chapter 3-Exercise Problems and Solutions
Solutions of the Relevant Problems
Chapter 3: PHY209-Fall2020
1
Phy209-Chapter 3-Exercise Problems and Solutions
3.7 Relevant Problems
Problem 3.1
Find the wavefunction and energy of the first excited state of the simple harmonic oscillator.
Solution
We know that the wavefunction and energy of the simple harmonic oscillator in the n-th excited state
are respectively as
1 n
ψn ( x) =
aˆ + ψ 0 ( x ) ,
n!
and
1

E n =  n +  hω ,
2

where
1
1 
d

aˆ + =
(−ipˆ + mω xˆ ) =
 − h + mω x  ,
dx
2hmω
2hmω 

and the ground state wavefunction is
1/ 4
 mω x 2 
 mω 
.

ψ 0 ( x ) = 
exp −


2
h
 πh 


Thus, for first excited state, n = 1
and
1 1
ψ1 ( x ) =
aˆ + ψ 0 ( x ) = aˆ + ψ 0 ( x ) .
1!
Thus, we have
ψ1 ( x ) =
d

  mω 

 − h + mω x  
dx
2 hmω 
 πh 
1
1/4
 mω x 2 
 .
exp −


2
h


After simplification, we get
1/4
 4 m3 ω3 
 mω x 2 


ψ1 ( x ) = 
exp −
 π h3 


2
h




as the wavefunction and
1
3

E1 =  1 +  hω = hω
2
2


as the energy of the first excited state of the simple harmonic oscillator.
Problem 3.2
Find the expectation value of the potential energy of the SHO in the n-th excited state.
Solution
The expectation value of the potential energy in the n-th excited state is given by
1
V n = mω2 x 2
2
n
1
= mω2 x 2
2
n
∞
1
= mω2  ψ n* xˆ 2 ψ n dx.
2
−∞
2
Phy209-Chapter 3-Exercise Problems and Solutions
Since, we know
1
aˆ ± =
2hmω
(m ipˆ + mω xˆ ) ,
we have
aˆ + + aˆ − =
2mω
xˆ ,
h
aˆ + − aˆ − =
1
2
pˆ ,
i hmω
and
from which we have
h
(aˆ + + aˆ − ),
2mω
xˆ =
and
hmω
(aˆ + − aˆ − ).
2
pˆ = i
Thus, we have
(
)
h
(aˆ + + aˆ − )(aˆ + + aˆ − ) = h aˆ +2 + aˆ + aˆ − + aˆ − aˆ + + aˆ 2− .
2mω
2mω
Therefore, the expectation value of the potential energy in the n-th excited state becomes
xˆ 2 =
mω2 h
V n=
ψn* aˆ +2 + aˆ + aˆ − + aˆ − aˆ + + aˆ −2 ψn dx.
2 2mω 
∞
(
)
−∞
But aˆ ±2 ψn  ψn ± 2 , apart from a normalization constant, which are orthogonal to ψ n . Therefore, these
terms in the above integral drop out, and we are left with
hω
V n=
4
=
=
∞
*
 ψ n (aˆ + aˆ − + aˆ − aˆ + ) ψn dx
−∞
∞
hω
4
hω
4
*
 ψn (aˆ + aˆ − ψn + aˆ − aˆ + ψn ) dx
−∞
∞
*
 ψn [nψn + (n + 1) ψn ] dx
−∞
∞
=
hω
(2n + 1)  ψ n* ψ n dx
4
−∞
hω 
1
n +  ,
2 
2
where, we have assumed that the state function ψ n is normalized.
=
3
Phy209-Chapter 3-Exercise Problems and Solutions
Problem 3.3
Show that the wavefunctions ψ m and ψ n of the m-th and n-th excited states of the particle in an
infinite one-dimensional potential well are orthogonal to each other for m ≠ n .
Solution
We know that the wavefunction of the m-th excited state of the particle in the infinite one-dimensional
potential well as
2  mπ x 
ψm ( x) =
sin
.
a  a 
Thus, we have
a
*
(ψ m ( x ), ψ n ( x )) = ∫ ψ m
( x )ψ n ( x ) dx
0
a
=
2
 mπ x   nπ x 
sin
 sin 
 dx
a∫  a   a 
0
a
=
1  π

π

cos (m − n) x  cos (m + n) x   dx

∫
a  a

a

0
=
 π
 π
1
1


sin (m − n) x  +
sin a (m + n) x  
(m − n)π   a
(
m
+
n
)
π
 0
 0
 
a
a
= 0 for m ≠ n.
Thus, we see that the inner product between ψ m and ψ n is zero for m ≠ n . Therefore, the
wavefunctions are orthogonal to each other.
Problem 3.4
For a single-step potential barrier, for E < V0 , in the region x > 0, the wavefunction of a particle is
given by ψ II ( x ) = G exp(−γ x ) . Show that the uncertainty in finding the particle in the region x > 0 is
given by Δx ≈ 1 / γ.
Solution
1/2
2
We know the uncertainty in x is given by Δx =  x 2 − x  .


We have
∞
*
 ψ II ( x) x ψ II ( x) dx
2
∞
1
 x exp( −2γ x) dx
∞
 y exp( − y) dy
14 γ 2 0
x =
=
=
∞
1
*
 ψ II ( x) ψ II ( x) dx G  exp(−2γ x) dx 14 γ  exp(− y) dy
0
∞
G
0
∞
2
0
0
0
1
Γ (2)
1
4γ 2
=
= .
1
2γ
Γ (1)
4γ
We also have
4
Phy209-Chapter 3-Exercise Problems and Solutions
∞
*
2
 ψ II ( x) x ψ II ( x) dx
x
2
=
0
∞

=
*
ψ II
( x ) ψ II ( x ) dx
0
2
∞
x
2
1
exp(−2γ x ) dx
Γ (3)
1
8γ 3
=
=
.
∞
2
1
γ
2
2
Γ (1)
G  exp(−2γ x ) dx
8γ
G
0
0
Therefore,
2
1/2  1
 1  
2
Δx =  x 2 − x 
=
−   


 2γ 2  2γ  


from which we can say that Δx ≈ 1 / γ.
1/2
 1 
=

2
 2γ 
1/2
=
1
2γ
,
Problem 3.5
An alpha particle is trapped in a nucleus of radius 1.4 × 10 −15 m. What is the probability that it will
escape from the nucleus if its energy is 2MeV? The potential barrier at the surface of the nucleus is
4MeV and the mass of the α − particle is 6.64 × 10 −27 kg.
Solution
We know transmission probability is
T = 16
E
V0

E
 1 −
V0


 2a

 exp −
2m(V0 − E )  .
 h


We have
(
)
2m(V0 − E ) = 2(6.64 × 10 − 27 kg)(4 − 2 )× 10 6 eV × 1.6 × 10 − 19 J/eV = 6.52 × 10 − 20 kg m s -1 .
Thus,
2a
2(2.8 × 10 −15 m)
2m(V0 − E ) =
× 6.52 × 10 − 20 kg m s -1 = 3.477.
−
34
h
1.05 × 10
Js
(
)
Therefore,
2
 2 
T = (16)  1 −  exp( −3.477) = 0.124.
4
 4 
(Answer)
Problem 3.6
Find x and px for the ground state of the linear harmonic oscillator.
Solution
We know the ground state wavefunction of the simple harmonic oscillator is
1/ 4
 mωx 2 
 mω 
ψ0 ( x) = 
exp−

,
2h 
 πh 

where symbols have usual meaning.
We have
1/2 ∞
∞
 mωx 2 
 mω 
x =  ψ 0 * ( x ) x dx = 
x exp−

 dx = 0.

πh 
h




−∞
−∞
Similarly,
∞
px =

d 
 ψ0 * ( x)  − ih dx  ψ0 ( x)dx == 0.
−∞
5
(p3.6a)
(p3.6b)
(p3.6c)
Phy209-Chapter 3-Exercise Problems and Solutions
Problem 3.7
A harmonic oscillator is in the ground state, (i) Where is the probability density maximum?
(ii) What is the value of maximum probability density?
Solution
We know the ground state wavefunction of the simple harmonic oscillator is
1/4
 mωx 2 
 mω 
exp−
ψ0 ( x) = 
(p3.7a)

,
2h 
 πh 

where symbols have usual meaning. We have the probability density in the ground state
1/2
 mωx 2 
2  mω 
(p3.7b)
exp−
ρ0 ( x ) = ψ 0 ( x ) = 

.
h 
 πh 

(i) For maximum probability density, we must have
1/2
 mωx 2   mω 
dρ0 ( x )
 mω 
=0  
(p3.7c)
 exp −
 −
 2 x = 0  x = 0.
dx
h   h 
 πh 

(ii) The probability density at x = 0 is
 mω 
ρ 0 ( x = 0) = 

 πh 
1/2
 mω 
exp[0] = 

 πh 
1/2
.
Problem 3.8
For the ground state of the linear harmonic oscillator, evaluate the uncertainty product ( Δx ) ( Δpx ).
Solution
According to the Virial theorem, the average values of the kinetic and potential energies of a classical
harmonic oscillator are equal. Assuming that this holds for the expectation values of the quantum
oscillator, we have the average values of the kinetic energy and the average potential energy in the n-th excited
state of the simple harmonic oscillator
T = V =
hω 
1
n + 
2 
2
or
1
1
1
hω 
p2x = mω2 x 2 =
 n + .
2m
2
2 
2
Therefore, we can write
h 
1
x2 =
n + 
mω 
2
and
1

p2x = mhω n +  .
2

But, we know the uncertainty Δx in x and uncertainty Δp x in p x are given by
h 
1
h 
1
2
( Δx ) 2 = x 2 − x =
n + − 0 =
n + 
mω 
2
mω 
2
and
1
1


2
( Δp x )2 = p 2x − p x = mhω n +  − 0 = mhω n + .
2
2


Therefore,
1

( Δx )( Δp x ) = h n +  .
2

Now setting n = 0 for the ground state, we have ( Δx )( Δp x ) = h / 2 .
6
(p3.8a)
(p3.8b)
(p3.8c)
(p3.8d)
(p3.8e)
(p3.8f)
Phy209-Chapter 3-Exercise Problems and Solutions
Problem 3.9
For a particle in a one-dimensional single step potential barrier, in the region x > 0, and when the
energy of the particle is smaller than the potential barrier height, the wavefunction of the particle is
ψ1 ( x ) = exp( − k 1 x ). Show that the position uncertainty Δx ≈ 1 / k1 .
Solution
First we normalize the given wavefunction ψ1 ( x ) = exp(−k1 x ). Let the normalized wavefunction be
ψ + ( x) = A exp(−k1 x ), where A is the normalization constant.
Applying the normalization condition
∞
 ψ + ( x)
2
dx = 1 which gives
A = 2k 1 .
(p3.9a)
0
Thus the normalized wavefunction is
ψ + ( x ) = 2k1 exp( − k1 x).
We know the position uncertainty can be written as
x2 − x
Δx =
2
(p3.9b)
(p3.9c)
.
We have
∞
x =  ψ +* xψ + dx 
x =
0
1
,
2k1
(p3.9d)
and
∞
x 2 =  ψ +* x 2ψ + dx 
x2 =
0
1
2k12
.
(p3.9e)
Using the expressions for x and x 2 from above to Eq. (p3.9c), we obtain
Δx =
1
1
≈ .
2k 1 k 1
Problem 3.10
The time-independent wave function of a particle of mass m moving in a potential V ( x ) = α 2 x 2 ,
where α is a constant is

mα 2 2 
ψ ( x ) = exp  −
x .
2


h
2


Find the energy of the system.
Solution
We are given the wavefunction

mα 2 2 
ψ ( x ) = exp  −
x .
2


2
h


(p3.10a)
The second derivative of ψ(x) with respect x is
d 2ψ
dx 2
=−

2mα 2 
2mα 2 2 
2mα 2 2 
1−
x exp −
x .



h 2 
h2
h2



7
(p3.10b)
Phy209-Chapter 3-Exercise Problems and Solutions
Substituting this into the time-independent Schrödinger equation
−
we obtain
h 2 d 2ψ
+ Vψ = Eψ,
2m dx 2
(p3.10c)


h 2  2mα 2 
2mα 2 2   2 2
−
1−
x
+ a x = E.
2m 

h 2 
h2


After simplification, we finally get
(p3.10d)
−
E=
hα
2m
.
Problem 3.11
Consider the wavefunction
 x2 
 exp(ik x − iω t ) ,
 a2 


where A is a real constant. (i) Find the value of A; (ii) Calculate the expectation value
ψ ( x) = A exp −
p for this
wavefunction.
Solution
(i) Applying the normalization condition
∞
 ψ( x)
2
dx = 1,
−∞
we obtain
 2x 2 
exp
  − a2  dx = 1,


−∞
From which after evaluating the integral, we obtain
A2
∞
(p3.11a)
2
a.
π
(ii) We have the expectation value
(p3.11b)
A=
∞
p =


d 

 ψ *  − ih dx  ψ
−∞
= ( −ih ) A 2
dx
 x 2  − ikx + iω t  2 x
 x2

e
 −
exp −
+ ikx  exp −
 a2 
 a2
 a2

−∞



∞

 ikx − iω t
e
dx


(p3.11c)
∞
2
 2x2 

 2  2∞
 x dx + (−ih )(ik ) A 2  exp − 2 x  dx.
 A  exp −
= ( −ih ) −
 a2 
 a2 
 a 2  −∞
−∞




The first integral is zero because the function under the integral sign is an odd function.
Furthermore, by utilizing Eq. (p3.11a), we obtain
p = hk .
8
Phy209-Chapter 3-Exercise Problems and Solutions
Problem 3.12
For a particle of mass m, Schrödinger initially arrived at the wave equation
1 ∂ 2ψ
∂ 2ψ m2c 2
−
ψ.
c 2 ∂t 2 ∂x 2
h2
Show that a plane wave solution of this equation is consistent with the relativistic energy momentum
relationship.
=
Solution
For plane waves, the solution may be taken as
ψ( x, t ) = A exp[i(kx − ωt )].
Substituting this solution in the given wave equation, we obtain
(−iω)2
c2
or
− ω2
ψ = (ik )2 ψ −
= −k 2 ψ −
m2 c 2
h2
(p3.12a)
ψ
m2 c 2
.
(p3.12b)
c2
h2
Multiplying both sides of Eq. (p3.12b) by c 2 h 2 and writing hω = E and hk = p, we obtain
E 2 = c 2 p 2 + m2 c 4 ,
which is the relativistic energy-momentum relationship.
(p3.12c)
Problem 3.13
Using the time-independent Schrödinger equation, find the potential V(x) and energy E for which the
wave function
n
 x 
ψ( x ) =   e − x / x 0 ,
 x0 
where n and x0 are constants, is an eigenfunction. You may assume V ( x) → 0 as x → ∞ .
Solution
Differentiating the wavefunction ψ(x) with respect to x, we obtain
dψ n  x 
 
=
dx x 0  x 0 
n −1
e
− x / x0
n
1  x 
−   e − x / x 0
x0  x0 
(p3.13a)
and
d 2ψ
dx 2
=
n(n − 1)  x 
 
x 02  x 0 
n −2
e− x / x0 −
2n  x

x 02  x 0



n −1
e− x / x0 +
n
 n(n − 1) 2n
1   x  − x / x0
=
−
+   e
x 0 x x 02   x 0 
 x 2
 n(n − 1) 2n
1 
=
−
+  ψ ( x ).
2
x 0 x x 02 
 x
Substituting (p3.13b) into the time-independent Schrödinger equation
−
we obtain
h 2 d 2ψ
+ Vψ = Eψ ,
2m dx 2
n
1  x  − x / x0
  e
x 02  x 0 
(p3.13b)
(p3.13c)
9
Phy209-Chapter 3-Exercise Problems and Solutions
−
h 2  n(n − 1) 2n
−
+

2m  x 2
x0 x
1 
 ψ + Vψ = Eψ ,
x 02 
(p3.13d)
which gives
E − V (x) = −
h 2  n(n − 1) 2n
−
+

2m  x 2
x0 x
1 
.
x 02 
(p3.13e)
Since we are given V ( x) → 0 as x → ∞, we have
h2
(p3.13f)
.
2mx02
Thus, the potential energy becomes
h 2  n(n − 1) 2n 
V (x) =
−

.
2m  x 2
x0 x 
E=−
(3.13g)
Problem 3.14
The wave function of a one-dimensional system is
ψ ( x) = Ax ne − x / a ,
where A, n and a are constants.
If ψ(x) is an eigenfunction of the Schrödinger equation, find the condition on V(x) for the energy
eigenvalue E = −h 2 /(2ma2 ) . Also find the value of V(x).
Solution
We are given
ψ ( x ) = Ax ne − x / a .
(p3.14a)
dψ
A
= Anx n −1e − x / a − x n e − x / a ,
dx
a
(p3.14b)
Therefore,
and
d 2ψ

2n
xn 
= Ae − x / a n(n − 1) x n − 2 − x n −1 +  .
a
dx 2
a2 

Thus, the time-independent Schrödinger equation
−
becomes
−
h 2 d 2ψ
+ Vψ = Eψ ,
2m dx 2
(p3.14c)
(p3.14d)

h2
2n
xn 
Ae − x / a n(n − 1) x n − 2 − x n −1 +  + V ( x ) Ax ne − x / a = EAx ne − x / a ,
2m
a
a2 

(p3.14e)
which gives
h 2  n(n − 1) 2n 1 
− +  = E − V ( x ).
2m  x 2
ax a2 
From the above equation, it is obvious that V ( x) → 0 as x → ∞, thus, we have
−
h2
.
2ma2
Thus, the potential energy becomes
E=−
V ( x) = −
h2
2ma2
−
(p3.14f)
(p3.14g)
h 2  n(n − 1) 2n 1 
−
+ .

2m  x 2
a a2 
10
(3.14h)
Phy209-Chapter 3-Exercise Problems and Solutions
Problem 3.15
A stream of particles of mass m and energy E move towards a single-step potential barrier. If the
energy of particles E < V0 , show that there is a finite probability of finding the particles in the region
x > 0. Also, determine the flux of (i) incident particles, (ii) reflected particles, and (iii) the particles in
region II ( x > 0). Comment on the results.
Solution
The Schrödinger equation and its solution in the two regions are
d 2ψ I
dx
2
d 2 ψ II
dx
2
+ k 02 ψ I = 0,
k 02 =
− γ 2ψ II = 0,
2mE
h2
γ2 =
,
(p3.15a)
x<0
2m(V0 − E )
h2
,
x > 0.
(p3.15b)
ψ I = e ik 0 x + Be −ik 0 x ,
x<0
(p3.15c)
ψ II = Ce − γx ,
x > 0.
(p3.15d)
γx
The solution e in region II is left out as it diverges and the region is an extended one. The
continuity condition at x = 0 gives
1 + B = C,
ik 0(1 − B) = −γC.
(p3.15e)
Solving, we get
2ik 0
ik 0 + γ
(p3.15f)
,
C=
.
ik 0 − γ
ik 0 − γ
The Reflection coefficient is
2  ik + γ   − ik 0 + γ 
 
 = 1
(p3.15g)
R = B =  0
 ik 0 − γ   − ik 0 − γ 
The Reflected flux is
hk
hk
2
2
(p3.15h)
FR = − v R B = − 0 B = − 0 .
m
m
The negative sign indicates that it is from right to left. Since ψ II is real, the transmitted flux is
FR = 0 and, therefore, the transmission coefficient T = 0. However, the wavefunction in the region x
> 0 is given by
 2ik 0  − γx
 e ,
(p3.15i)
ψ II = 
x > 0.
 ik 0 − γ 
Therefore, the probability that the particle is found in the region x > 0 is finite. Due to the uncertainty
in energy, the total energy may even be above V0 .
B=
11
Phy209-Chapter 3-Exercise Problems and Solutions
Problem 3.16
Obtain the energy eigenvalues and eigenfunctions of a particle trapped in the potential V(x) = 0 for
0 < x < a and V ( x ) = ∞, otherwise. Show that the wave functions for the different energy levels of the
particle trapped in the square well are orthogonal.
Solution
From the theory (Fig. 3.2), the normalized wavefunctions are given by
2
 nπx 
ψn ( x) =
sin
.
a
 a 
Let us consider the wavefunctions ψ m (x ) and ψn (x ) . The inner product of them is
a
(p3.16a)
a
(ψ m ( x ), ψn ( x )) =  ψ m * ( x ) ψn ( x ) dx = 2  sin mπx  sin nπx 
a
0
 a 
0
 a 
dx
a
=
1   πx

 πx

cos (m − n) − cos (m + n)  dx

a  a


a
0
a
(p3.16b)
a
1
1
 πx

 πx

sin  (m − n) −
sin  (m + n)
π (m − n)  a
 0 π (m + n )  a
0
1, m = n
= δ mn = 
0, m ≠ n
From which it is clear that in m ≠ n , then the inner product of the wavefunctions ψ m ( x ) and ψ n ( x )
is zero and the wavefunctions are orthogonal to each other.
=
Problem 3.17
Obtain the energy eigenvalues and eigenfunctions of a particle trapped in the one-dimensional
potential box V(x) = 0 for 0 ≤ x ≤ a and V ( x ) = ∞, otherwise. Calculate the expectation values of
position < x > and of the momentum < px > of the particle trapped in the box.
Solution
From the theory (Fig. 3.2), the normalized wavefunctions are given by
2
 nπx 
ψn ( x) =
sin
.
a
 a 
Thus the average value of x is
a
x =  ψ n * ( x ) x ψ n ( x ) dx =
0
=
a
a
0
0
a
a
0
0
(p3.17a)
2
1 
 nπx 
 2nπx  
sin 2 
 x dx =  1 − cos
  x dx

a
a 
 a 
 a 
1
1
 2nπx 
x dx −  cos
 x dx

a
a
 a 
a
= .
2
Thus the average value of p x is
a
a
d 
2

 nπx   nπ 
 nπ x 
p x =  ψ n * ( x ) − ih  ψ n ( x ) dx = (−ih )  sin 
   cos
 dx
dx 
a

 a  a 
 a 
0
0
a
 nπ  1
 2nπx 
= ( −ih )   sin 
 dx
 a  a0  a 
= 0.
12
Phy209-Chapter 3-Exercise Problems and Solutions
Problem 3.18
An electron in a one-dimensional infinite potential well, defined by V(x) = 0 for 0 ≤ x ≤ a and
V ( x) = ∞, otherwise, goes from the n = 4 to the n = 2 level. The frequency of the emitted photon is
3.43 × 1014 Hz. Find the width of the box.
Solution
From the theory (Fig. 3.2), the energy eigenvalues are given by
En =
π 2 n2 h 2
2ma2
From Eq. (3.18a), we have
E 4 − E2 =
16π 2 h 2
2ma2
From which we obtain
−
.
(p3.18a)
4 π 2h 2
2ma2
=
6π 2h 2
ma2
= hf ,
(p3.18b)
6 π 2h 2
3h
(3)(6.64 × 10 −34 J.s)
=
=
= 3.19 × 10 −18 m 2 , (p3.18c)
mhf
2mf (2)(9.1 × 10 − 31 kg)(3.43 × 1014 Hz)
from which we obtain
a2 =
a = 1.79 × 10 −9 m.
Problem 3.19
A particle of mass m trapped in the potential V(x) = 0 for 0 ≤ x ≤ a and V ( x ) = ∞, otherwise. Evaluate
the probability of finding the trapped particle between x = 0 and x = a/n when it is in the n-th excited
state.
Solution
In the above situation, we know the wavefunction of the particle is
ψn ( x) =
2
 nπx 
sin
.
a
 a 
(p3.19a)
Therefore, the probability density is
2
ρn ( x ) = ψ n ( x ) =
2
 n πx 
sin 2 
.
a
 a 
(p3.19b)
The required probability is
a/n
Pn =
 ρn ( x) dx =
0
2
a
a/n
a/n
1
2  nπx 
 sin  a  dx = a 
0
0
13

1
 2nπx  
1 − cos a   dx = n .



Phy209-Chapter 3-Exercise Problems and Solutions
Problem 3.20
The wave function of a particle confined in a box of width a is
2  πx 
sin  ,
0 ≤ x ≤ a.
a  a 
Calculate the probability of finding the particle in the region 0 < x < a/2.
ψ1 ( x ) =
(p3.20a)
Solution
The solution can be obtained similar to problem 3.19. We have the probability density
2
ρ1 ( x ) = ψ1 ( x ) =
2
 πx 
sin 2   .
a
 a 
(p3.20b)
The required probability is
a/2
P=
a/2
a/2

1
1
 πx 
 2 πx 
sin 2   dx =
1 − cos
 dx = .


a
2
 a 
 a 
0
0 
2
 ρ1 ( x) dx = a 
0
Problem 3.21
Consider a particle of mass m moving in a one-dimensional potential specified by
 0, − 2a < x < 2a
V(x ) = 
∞ , otherwise.
Find the energy eigenfunctions and eigenvalues.
Solution
Similar to the theory of Sec. 3.3.
Problem 3.22
For an electron in a one-dimensional infinite potential well of width 1 Å, calculate (i) the separation
between the two lowest energy levels; (ii) the frequency and wavelength of the photon corresponding
to a transition between these two levels; and (iii) in what region of the electromagnetic spectrum is
this frequency/wavelength?
Solution
We assume that the one-dimensional infinite potential well is defined by V(x) = 0 for 0 ≤ x ≤ a and
V ( x) = ∞, otherwise.
From the theory of Sec. 3.3, (Fig. 3.2), the energy eigenvalues of the electron are given by
En =
π 2 n2 h 2
2ma2
From Eq. (3.22a), we have
E2 − E1 =
4 π 2h 2
2ma2
(i) We have
E2 − E1 =
3π 2 h 2
2
−
.
(p3.22a)
π 2h 2
2ma2
=
3h 2
2
3π 2 h 2
=
2ma2
=
= hf .
(3)(6.63 × 10 −34 J.s)2
2ma
8ma
(8)(9.1 × 10
(ii) From Eq. (p3.22b), we have
f=
3π 2 h 2
2
2ma h
=
3h
8ma
2
=
(p3.22b)
- 31
kg)(10
−10
m)
2
= 1.8 × 10 −17 J.
(3)(6.63 × 10 −34 J.s)
(8)(9.1 × 10
- 31
14
kg)(10
− 10
m)
2
= 2.73 × 1016 s -1 .
Phy209-Chapter 3-Exercise Problems and Solutions
The wavelength is
c
3 × 10 8 ms -1
=
= 1.1 × 10 − 8 m.
f 2.73 × 1016 s -1
(iii) This frequency falls in the vacuum ultraviolet region.
λ=
Problem 3.23
A 1eV electron got trapped inside the surface of a metal. If the potential barrier is 4.0eV and the width
of the barrier is 2Å, calculate the probability of its transmission.
Solution
If a is the width of the barrier, then the transmission coefficient is
E
E
 2a

T = 16  1 −  exp  −
2m(V − E )  .
(p3.23a)
V
V
 h

Substituting the given values of the parameters in the above expression for the transmission
coefficient, we obtain
 (2)(2 × 10 −10 m)

1 1
T = 16  1 −  exp−
(2)(9.1 × 10 − 31 kg)(3 × 1.6 × 10 −19 J)  = 0.085.
− 34
4 4
J.s.
 1.05 × 10

Problem 3.24
A beam of 12eV electrons is incident on a potential barrier of height 30eV and width 0.05nm.
Calculate the transmission coefficient.
Solution
The solution is similar to the solution of Problem 3.23.
Problem 3.25
A beam of particles having energy 2eV is incident on a potential barrier of 0.1 nm width and 10eV
height. Show that the electron beam has a probability of 14% to tunnel through the barrier.
Solution
The expression for the transmission probability is
16 E(V0 − E)
2m(V0 − E)
T=
exp[− 2γa],
γ2 =
,
(p3.25a)
2
V0
h2
where a is the width of the barrier, V0 is the height of the barrier, and E is the energy of the electron.
We have
γ2 =
(2)(9.1 × 10 −31 kg)(8 × 1.6 × 10 −19 J)
(1.05 × 10
from which, we obtain
γ = 1.45 × 10 10 m -1 .
Therefore,
− 34
J.s.)
2
= 2.113 × 10 20 m -2 ,
(p3.25b)
(p3.25c)
γa = (1.45 × 1010 m -1 )(0.1 × 10 -9 m) = 1.4536.
Thus the transmission probability is
(16)(2 eV )(8 eV )
T=
exp[− (2)(1.4536] = 0.14 = 14%.
(10 eV )2
15
(p3.25d)
Phy209-Chapter 3-Exercise Problems and Solutions
Problem 3.26
An electron having energy E = 1eV is incident upon a rectangular barrier of potential energy
V0 = 2 eV. How wide must the barrier be so that the transmission probability is 10 −3 ?
Solution
The expression for the transmission probability is
E 
E 
 2a

 1 −
 exp −
(p3.26a)
2m(V0 − E )  ,
T = 16
V0  V0 
 h

From the above equation, we can write an expression for the width of the barrier as
 16 E 
0.5h
E 
 1 −
 .
ln 
(p3.26b)
a=
2m(V0 − E )  T V0  V0  
Substituting the values of different parameters in the above expression, we obtain
a=
1.05 × 10 −34 J.s.
 16  1  1  
  1 −   = 1.67 × 10 − 9 m.
ln 
− 3  2 
− 31
−19
2 
(2)(9.1 × 10 kg)(1 × 1.6 × 10 J )  10 
****************************
16