CH153 Quantum Mechanics: Workshop 2
Worked Solutions
Dr. T. R. Walsh
1. A neutron is trapped in a 1D box, with box length L = 1.6 Å. The potential inside the
box is zero. The kinetic energy of the neutron is 7.2x10−2 eV.
a) What is the momentum of the neutron, in units of kg m s−1 ?
Answer: You will need to find out the mass of a neutron (if needed in an exam, this would
be provided). From P. W. Atkins’ “Physical Chemistry”, m = 1.67492x10−27 kg. The kinetic
energy is given, so it’s fastest to use the relation between KE and momentum (p):
KE =
p2
.
2m
Rearrange this so that p2 is on the left-hand-side (l.h.s):
p2 = 2m × KE
and then take the square-root of both sides to get p:
√
p = 2m × KE.
Remember that when you substitute your values for the mass and KE in this express, you use
mass in kg and KE in Joule. So you will have to convert your KE. Use the conversion factor
−2
−19
−20
1 eV = 1.602×10−19 J. Therefore, 7.2x10−2 eV =
√ 7.2x10 × 1.602×10 J = 1.153×10 J.
That is, the momentum of the neutron is p = 2 × 1.67492 × 10−27 × 1.153 × 10−20 .
So, p =6.2×10−24 kg m s−1 .
End of answer to 1a)
b) Calculate the de Broglie wavelength of the neutron at this kinetic energy.
Answer: Now that you have the momentum of the neutron, just make use of the de Broglie
relation:
h
p
6.626 × 10−34
=
6.2 × 10−24
= 1.07Å
λB =
End of answer to 1b)
1
c) What value of n (our principal quantum number for the particle in a box) does the neutron
kinetic energy correspond to? How many nodes will the corresponding wavefunction have?
Answer: First, you should remember that total energy is conserved, i.e. E = KE + V. You
should also realise that in a 1D box where the potential (V) is zero inside the box, the total
energy, E, corresponds to the KE. Make use of the expression you have for the energy of a
particle in a 1D box:
n2 h2
En =
.
8 m L2
Since KE = total energy, then KE = En . Since the KE is given, you need to rearrange this
equation to get n2 on the l.h.s:
8 m L2 × KE
n2 =
.
h2
Now just plug in your values for mass, box length (L), KE and h. Don’t forget to square
your box length: L = 1.6 Å= 1.6 ×10−10 , so L2 = 2.56×10−20 . Again, you have to use KE
in units of joule.
8 × (1.67492 × 10−27 ) × (2.56 × 10−20 ) × (1.153 × 10−20 )
(6.626 × 10−34 )2
≈ 9
n2 =
So, taking the square-root of both sides yields n ≈ 3. Since there are (n − 1) nodes in the
wavefunction Ψn , and n=3 in our solution, this means the corresponding wavefunction will
have 2 nodes.
End of answer to 1c)
d) Do you expect the de Broglie wavelength to increase or decrease when n (quantum
number) is increased? (Hint: consider the correspondence principle).
Answer: When our quantum number n increases, classical behaviour becomes more apparent—
this is the correspondence principle. Since you know from the previous workshop that the
de Broglie wavelength is very small for ‘classical’, macroscopic objects, expect that as n
increases, λB will decrease.
End of answer to 1d)
2. A research student has obtained (using a computer) some wavefunctions for a molecular
system. There must be a bug in the computer program, as some of these wavefunctions are
unacceptable solutions under the Born interpretation.
Which of a), b) and c) are unacceptable solutions? For each unacceptable solution, indicate
what errors are present.
Answer: Remember the 4 criteria by which we judge acceptable solutions to the Schrödinger
equation, under the Born interpretation: Ψ must not be infinite anywhere, Ψ must be
single-valued everywhere (only one value of the wavefunction for each x), Ψ must not be
discontinuous, and Ψ must have a continuous gradient, and also have finite 2nd derivatives
at every point.
2
So for part (a), Ψ is not acceptable, since this solution is discontinuous. For part
(b), Ψ is an acceptable solution. For part (c), Ψ is not acceptable, since this solution is
many-valued (i.e. not single-valued).
End of answer to 2
3. Consider the particle in a 1D box system, where n = 3. The wavefunction is given by
s
Ψ3 (x) =
2
3πx
sin(
).
L
L
Recall from your notes that the probability of finding the particle, inside the box, between
x=l1 and x=l2 , for Ψn (x) is
Pl1 ,l2 =
l2 − l1
1
2nπl1
2nπl2
+(
)[sin(
) − sin(
)]
L
2nπ
L
L
where l2 > l1 .
a) Sketch Ψ3 (x) and Ψ23 (x). How many nodes does Ψ3 (x) have?
Answer: In the following graphs, B =
q
2/L.
3
The graph of [Ψ3 (x)]2 has been divided into 5 equal segments. Ψ3 (x) has 2 nodes.
End of answer to 3a)
b) Just we have done in the lectures, divide the 1D box into 5 equal portions, with intervals
from x=0 to x= L5 , x= L5 to x= 2L
, x= 2L
to x= 3L
, x= 3L
to x= 4L
,x= 4L
to x=L. Determine
5
5
5
5
5
5
the probability Pl1 ,l2 for each of these 5 intervals. (Do you need to do the calculation for
every interval?) Can you rationalise your results by inspection of your sketch of Ψ23 (x)?.
Answer: Looking at the first segment (x=0 → x = L/5). l2 = L/5 and l1 = 0. Remember,
1
1
n=3. The angle (6π/5) radians is 216◦ . The prefactor 6π
must be evaluated as 6×3.1415926...
=
1
= 0.0530516 . . ..
18.849556...
1
6π × 0
6π(L/5)
)[sin(
) − sin(
)]
6π
L
L
6π
1
0.2 + ( )[0 − sin( )]
6π
5
1
0.2 + ( )[0 − sin(216◦ )]
6π
0.2 − −0.0312
0.231
P0,L/5 = 1/5 + (
=
=
=
=
Now consider the second segment (x = L/5 → x = 2L/5). l2 = 2L/5 and l1 = L/5. The
angle (12π/5) radians is 72◦ .
PL/5,2L/5 = 1/5 + (
1
6π(L/5)
6π(2L/5)
)[sin(
) − sin(
)]
6π
L
L
4
6π
12π
1
)[sin( ) − sin(
)]
6π
5
5
1
= 0.2 + ( )[sin(216◦ ) − sin(72◦ )]
6π
= 0.2 − 0.082
= 0.118
= 0.2 + (
Now consider the third segment (x = 2L/5 → x = 3L/5). l2 = 3L/5 and l1 = 2L/5. The
angle (18π/5) radians is 288◦ .
1
6π(2L/5)
6π(3L/5)
)[sin(
) − sin(
)]
6π
L
L
1
12π
18π
0.2 + ( )[sin(
) − sin(
)]
6π
5
5
1
0.2 + ( )[sin(72◦ ) − sin(288◦ )]
6π
0.2 + 0.101
0.301
P2L/5,3L/5 = 1/5 + (
=
=
=
=
Look at the red area under the curve; the area under the curve for the region x=4L/5 →
x=L is the same as the area under the curve for the region x=0 → x=L/5. This is just by
symmetry of the function. So, P4L/5,L = P0,L/5 .
Look at the green area under the curve; the area under the curve for the region x=3L/5 →
x=4L/5 is the same as the area under the curve for the region x=L/5 → x=2L/5. This is
just by symmetry of the function. So, P3L/5,4L/5 = PL/5,2L/5 .
Inspection of the area under each segment of the curve would lead you to expect the central
segment (x=2L/5 → x=3L/5) to have the largest area, followed by the red segments. The
green segments should have the smallest area—these regions contain the wavefunction nodes.
End of answer to 3b)
4. Consider a particle of mass, m, trapped in a 3D box. The box is a cube, with dimensions
Lx = Ly = Lz = d. The expression for the energy of the system now takes three quantum
numbers, nx , ny and nz , and is as follows
Enx ,ny ,nz =
n2
n2
h2 n2x
( 2 + y2 + z2 )
8 m Lx L y Lz
=
h2
(n2x + n2y + n2z )
2
8md
a) What values of nx , ny , nz correspond to the zero-point energy? Write down the expression
for the zero-point energy of this system, labelling it as Enx ,ny ,nz , with the appropriate values
of nx , ny , nz .
5
Answer: The zero-point energy is defined as the lowest possible energy that the system can
take: it must correspond to nx = 1, ny = 1, nz = 1. We express this as:
h2
(12 + 12 + 12 )
8 m d2
h2
=
(3)
8 m d2
3h2
=
8 m d2
= E111
Ezpe =
End of answer to 4a)
b) Identify all energy levels (i.e. all different combinations of (nx , ny , nz )) up to and including
5 times the zero-point energy. Label each energy level as Enx ,ny ,nz , and express each energy
h2
. How many unique energies are there?
as a multiple of 8 m
d2
Answer: First, you need to recognise that 5 times the zero-point energy corresponds to
5×
3 h2
15 h2
=
.
8 m d2
8 m d2
Then, all you need to do is find all unique combinations of (nx , ny , nz ) that amount to 15 or
less:
h2
(12 + 12 + 22 )
2
8md
h2
=
(6)
8 m d2
6h2
= E121 = E211
=
8 m d2
E112 =
h2
(12 + 22 + 22 )
8 m d2
h2
=
(9)
8 m d2
9 h2
=
= E212 = E221
8 m d2
E122 =
h2
(12 + 12 + 32 )
8 m d2
h2
=
(11)
8 m d2
11 h2
=
= E131 = E311
8 m d2
E113 =
6
h2
(22 + 22 + 22 )
8 m d2
h2
=
(12)
8 m d2
12 h2
=
8 m d2
E222 =
h2
(32 + 22 + 12 )
8 m d2
h2
=
(14)
8 m d2
14 h2
=
8 m d2
= E312 = E213 = E231 = E123 = E132
E321 =
There are six unique energies.
End of answer to 4b)
c) For each unique energy, write down how many energy levels (i.e.
(nx , ny , nz )) there are.
combinations of
2
[Hint: consider E = 8 6h
. The combinations of (nx , ny , nz ) that are possible are (2,1,1),
m d2
(1,2,1), and (1,1,2)]
When a unique energy can be described by more than one combination of
nx , ny , nz , this is called degeneracy.
Answer:
3 h2
8 m d2
6 h2
8 m d2
9 h2
8 m d2
11 h2
8 m d2
12 h2
8 m d2
14 h2
8 m d2
is non − degenerate
is 3 − fold degenerate
is 3 − fold degenerate
is 3 − fold degenerate
is non − degenerate
is 6 − fold degenerate
End of answer to 4c)
For those who like a challenge...
5. Again consider a particle of mass, m, trapped in a 3D box, this time with box dimensions
Lx = Ly = d and Lz = 2d.
7
a) Write down the general expression for the energy as a multiple of
Answer:
Enx ,ny ,nz
h2
.
8 m d2
h2 n2x n2y
n2
=
( 2 + 2 + z 2)
8m d
d
(2d)
2
h
n2z
2
2
=
(n
+
n
+
)
y
8 m d2 x
4
End of answer to 5a)
b) Write down the expression for the zero-point energy of this system.
Answer:Remember, the zero-point energy is defined as the lowest possible energy that the
system can take: it must correspond to nx = 1, ny = 1, nz = 1. We express this as:
h2
12
2
2
(1
+
1
+
)
8 m d2
4
5 h2
= ( )
4 8 m d2
= E111
Ezpe =
End of answer to 5b)
2
c) Write down all energy levels (combinations of nx , ny , nz ) up to and including 811h
. Label
m d2
h2
each energy level as Enx ,ny ,nz , and express each energy level as a multiple of 8 m d2 .
Answer:
h2
22
2
2
(1
+
1
+
)
8 m d2
4
h2
=
(1 + 1 + 1)
8 m d2
3 h2
=
8 m d2
E112 =
h2
32
2
2
(1
+
1
+
)
8 m d2
4
h2
9
=
(1 + 1 + )
2
8md
4
17 h2
= ( )
4 8 m d2
E113 =
h2
42
2
2
(1
+
1
+
)
8 m d2
4
h2
16
=
(1 + 1 + )
2
8md
4
6h2
=
8 m d2
E114 =
8
h2
52
2
2
(1
+
1
+
)
8 m d2
4
h2
25
=
(1 + 1 + )
2
8md
4
2
33 h
= ( )
4 8 m d2
E115 =
h2
62
2
2
(1
+
1
+
)
8 m d2
4
h2
36
=
(1 + 1 + )
2
8md
4
2
11h
=
8 m d2
E116 =
h2
12
2
2
(1
+
2
+
)
8 m d2
4
h2
1
(1 + 4 + )
=
2
8md
4
21 h2
= ( )
= E211
4 8 m d2
E121 =
h2
22
2
2
(2
+
1
+
)
8 m d2
4
h2
=
(4 + 1 + 1)
8 m d2
6 h2
=
= E122
8 m d2
E212 =
h2
32
2
2
(1
+
2
+
)
8 m d2
4
h2
9
=
(1 + 4 + )
2
8md
4
29 h2
= ( )
= E213
4 8 m d2
E123 =
h2
42
2
2
(1
+
2
+
)
8 m d2
4
h2
16
=
(1 + 4 + )
2
8md
4
9h2
=
= E214
8 m d2
E124 =
9
h2
12
2
2
(2
+
2
+
)
8 m d2
4
h2
1
=
(4 + 4 + )
2
8md
4
33 h2
= ( )
= E115
4 8 m d2
E221 =
h2
22
2
2
(2
+
2
+
)
8 m d2
4
h2
4
=
(4 + 4 + )
2
8md
4
9h2
= E124 = E214
=
8 m d2
E222 =
E131
h2
12
2
2
=
(1 + 3 + )
8 m d2
4
h2
1
(1 + 9 + )
=
8 m d2
4
41 h2
= ( )
= E311 = E223
4 8 m d2
h2
22
2
2
(3
+
1
+
)
8 m d2
4
h2
=
(9 + 1 + 1)
8 m d2
11 h2
=
= E132
8 m d2
E312 =
End of answer to 5c)
d) How many unique energies are there? What is the degeneracy of each unique energy?
i.e. how many combinations of nx , ny , nz are there for each unique energy? Are there any
energies that are 3-fold degenerate?
2
2
h
Answer: There are 10 unique energies up to and including 811h
. They are: ( 54 ) 8 m
,
m d2
d2
2
2
2
2
2
2
2
2
17
h
21
h
6h
29
h
33
h
9h
41
h
11h
( 4 ) 8 m d2 , ( 4 ) 8 m d2 , 8 m d2 , ( 4 ) 8 m d2 , ( 4 ) 8 m d2 , 8 m d2 , ( 4 ) 8 m d2 , 8 m d2 .
2
3h2
,
8 m d2
h
There are 3 unique energies that are non-degenerate – they are the following: ( 54 ) 8 m
(E111 ),
d2
2
2
3h
17
h
(E112 ), ( 4 ) 8 m d2 (E113 ).
8 m d2
10
2
There are 3 unique energies that are doubly-degenerate – they are the following: ( 21
) h
4 8 m d2
2
2
(E121 , E211 ), ( 29
) h (E213 , E123 ), ( 33
) h (E115 , E221 ).
4 8 m d2
4 8 m d2
2
There are 4 unique energies that have 3-fold degeneracy – they are the following: 8 6h
(E114 ,
m d2
9h2
41
h2
11h2
E212 E122 ), 8 m d2 (E222 , E214 E124 ), ( 4 ) 8 m d2 (E223 , E311 E131 ), 8 m d2 (E116 , E132 E312 ).
End of answer to 5d)
6. Consider an electron trapped in a 1D well, of finite depth, V = 4.0 eV. The well has
boundaries at x = 0 and x = L. The potential inside the well is zero. The electron has an
energy of 1.5 eV.
Outside the well boundaries, you know that the wavefunction will decay exponentially.
Consider the boundary at x = 0. The wavefunction will decay like
Ψ(x) ∼ eαx , x < 0.
where
s
α=
2 m (V − E)
h̄2
The mass of an electron is 9.109x10−31 kg.
What is the value of eαx when x = −1x10−9 ? (the unit of x is actually metres)
What is the value of eαx when x = −0.1x10−9 m?
What is the value of eαx when x = −10x10−9 m?
Do your results change if the width of the well is doubled? (everything else is kept fixed,
including the electron energy)
Answer: A schematic diagram for this 1D well is shown below.
The first thing to do in this problem is determine α. To calculate α, you need to find (V E) in joule:
V −E =
=
=
=
4 − 2.15
2.5 eV
2.5 × (1.602 × 10−19 J
4.005 × 10−19 J
11
Therefore,
α
v
u
u (4.005 × 10−19 ) × (9.109 × 10−31 ) × 2
= t
−34 2
√
(1.055 × 10
)
=
6.56 × 1019
= 8.1 × 109 m−1
So, when x = −1 nm (= −1×10−9 m), then
9
eαx = e(8.1×10 )×(−1×10
= e−8.1
= 3 × 10−4
−9 )
When, x = −0.1 nm (= −1×10−10 m), then
eαx = e(8.1×10
= e−0.81
= 0.44
9 )×(−1×10−10 )
When, x = −10 nm (= −1×10−8 m), then
9
eαx = e(8.1×10 )×(−1×10
= e−81
= 7 × 10−36
−8 )
There is no explicit dependence of the well width (L) on α. Therefore the results do not
change if the well width is doubled (and nothing else is changed).
End of answer to 6
7. Repeat question 6, this time considering a proton instead of an electron, also at an energy
of 1.5 eV. The mass of a proton is 1.67265x10−27 kg.
Compare your results with those obtained for the electron under the same conditions. Given
that Ψ(x)2 is proportional to the probability of finding a particle in a given region of space,
which tunnels detectably further into the well wall, the proton or the electron?
Answer: Your calculation for α from question 6 is repeated, this time using the mass of the
proton.
α
v
u
u (4.005 × 10−19 ) × (1.67265 × 10−27 ) × 2
= t
−34 2
√
(1.055 × 10
=
1.205 × 1033
= 3.47 × 1011 m−1
12
)
So, when x = −1 nm (= −1×10−9 m), then
eαx = e(3.47×10
= e−347
≈ 0
11 )×(−1×10−9 )
When, x = −0.1 nm (= −1×10−10 m), then
11
eαx = e(3.47×10 )×(−1×10
= e−34.7
= 8 × 10−16
−10 )
When, x = −10 nm (= −1×10−8 m), then
eαx = e(3.47×10
= e−3471
≈ 0
11 )×(−1×10−8 )
Since a larger value of α leads to faster decay of the wavefunction inside the potential barrier,
then it is the electron which will tunnel detectably further into the barrier. (Remember that
probability of finding the particle is related to Ψ2 under the Born interpretation).This is
evidenced by the numerical values of Ψ as a function of distance inside the barrier (shown
above in Q6 and Q7).
End of answer to 7
8. A particle of mass, m, strikes a potential barrier of height V = 6 eV. This particle can
have one of four energies: E1 = 0.05 eV,E2 = 1.2 eV, E3 = 3.6 eV, E4 = 4.5 eV.
The probability of finding the particle inside the barrier when it has energy E1 is denoted
P1 . P2 , P3 and P4 are similarly denoted.
You know that the wavefunction will decay exponentially inside the barrier, and that (in
1D) the probability of finding a particle in a given region of space is the integral of Ψ(x)2
over that space (i.e. the area under the curve for a given range of x). Without doing the
integral (i.e. by inspection), rank the probability of finding the particle inside the barrier
(i.e. rank P1 , P2 , P3 and P4 ) from smallest to greatest.
Answer: The barrier height is V = 6 eV. Therefore:
(V
(V
(V
(V
− E1 ) = 6.0 − 0.05 = 5.95 eV
− E2 ) = 6.0 − 1.20 = 4.80 eV
− E3 ) = 6.0 − 3.60 = 2.40 eV
− E4 ) = 6.0 − 4.50 = 1.50 eV
Recall that α is defined as
s
α=
2 m (V − E)
h̄2
13
and that the only thing changing here is the term (V - E). Therefore
s
2 m (V − E1 )
h̄2
s
2 m (V − E2 )
=
h̄2
s
2 m (V − E3 )
=
h̄2
s
2 m (V − E4 )
=
h̄2
α1 =
α2
α3
α4
It should be clear that α1 > α2 > α3 > α4 . Since Ψ(x) ∼ e−αx (x > 0), then the largest value
of α will yield the fastest decay inside the barrier, and consequently the ‘smallest’ tunnelling
(and also the smallest area under [Ψ(x)]2 ).
Therefore the chance of finding the particle in the ‘classically forbidden’ zone can be ranked:
P 1 < P 2 < P3 < P 4 .
End of answer to 8
9. A 1D potential well of unknown depth, V, contains an electron. The potential inside the
well is zero, the potential outside the well is V. The electron is in a bound state with energy
E = 0.14 eV. The wavefunction associated with this electron decays exponentially outside
the well walls in the usual way:
Ψ(x) ∼ e−αx
(x > 0).
The value of the decay constant (α) is α = 9.049×109 m−1 .
a) What is the depth, V, of this potential well, expressed in units of eV?
Answer: The key to solving this problem is to rearrange the expression
s
α=
2 m (V − E)
h̄2
so that you have (V - E) on the l.h.s. To do this, start by writing α2 :
2 m (V − E)
h̄2
2
α2 h̄
(V − E) =
2m
α2 =
therefore,
Now just substitute in the values for α, h̄ and the mass:
(9.049 × 109 )2 × (1.055 × 10−34 )2
2 × 9.109 × 10−31
(8.188 × 1018 ) × (1.112 × 10−68 )
=
2 × 9.109 × 10−31
= 4.999 × 10−19 J
(V − E) =
14
You are told in the question that E = 0.14 eV. Convert this into joule, in the usual way, i.e.,
E = 0.14 × 6.102×10−19 J = 2.243×10−20 J. It follows that
V
= 4.999 × 10−19 + E
= 4.999 × 10−19 + 2.243 × 10−20
= 5.223 × 10−19 J
Finally, you need to convert your answer for V from joule into eV: V = 5.223×10−19 J =
5.223×10−19 ÷ 1.602×10−19 eV = 3.26 eV.
End of answer to 9a
b) An incoming photon with frequency 1.088×1015 Hz strikes the electron. Consequently, the
electron escapes the potential well. What is the maximum possible velocity of the escaping
electron?
Answer: This is a photoelectric-effect type question. The first thing to do is determine the
energy of the incoming photon
Ephoton = hν
= 6.626 × 10−34 × 1.088 × 1015
= 7.21 × 10−19 J
(V − E) is the amount of energy it takes for the electron to escape the potential well. So,
recall that the maximum kinetic energy of the escaping electron will be equal to the energy
of the photon minus (V - E):
KEmax = hν − (V − E)
= 7.21 × 10−19 − 4.999 × 10−19
= 2.21 × 10−19 J
2
:
Now make use of the expression for the kinetic energy, KEmax = 12 mvmax
1 2
mv
2 max
2 × KEmax
=
m
2 × 2.21 × 10−19
=
9.109 × 10−31
= 4.85 × 1011 m2 s−2
KEmax =
2
so, vmax
2
Just take the square root of vmax
to find vmax = 6.97 × 105 m s−1 .
End of answer to 9b
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