PC 242 Assignment 4
1. Through what potential difference would an electron have to be accelerated to give it a
de Broglie wavelength equal to the diameter of the hydrogen atom? How does the kinetic
energy of this electron compare to the ground state energy of the hydrogen atom?
The radius of a hydrogen atom is the Bohr radius of 0.0529nm. The diameter is double
the Bohr radius. Setting this equal to the de Broglie wavelength of an electron, the
corresponding momentum of the electron is
h
h
p= =
λ 2a0
The corresponding kinetic energy would be
p2
h2
h2c2
(1240eV ⋅ nm )2
KE =
=
=
=
= 134.4eV
2m 8me a02 8me c 2 a02 8(0.511MeV )(0.0529nm)2
The voltage (potential difference) required to impart this energy is obtained by setting the
potential energy equal to this kinetic energy:
h2
h2
eV =
⇒
V
=
= 134.4V
8me a02
8me a02 e
The kinetic energy of the electron is a factor of 10 larger than the ground state energy of
the hydrogen atom.
2. In Davisson and Germer’s experiment, electrons were accelerated through a potential
difference of 50.0V. If the crystal spacing is known to be 2.15 Angstroms, at what angle
would the 3rd dark ring be observed?
The condition for destructive interference is that the path difference between waves is a
half integer number of wavelengths:
λ
d sin θ = n
2
For the 3rd dark ring, n=3. D is the crystal spacing. The de Broglie wavelength is related
to the momentum by
h
λ=
p
The momentum p can be calculated from the kinetic energy imparted to the electron by a
potential difference of 50.0 V:
p2
= eV ⇒ p = 2me eV
2me
Therefore
$
'
h
h
d sin θ = n
⇒ θ = sin −1 & n
)
2 2me eV
% 2d 2me eV (
Note that the value of the voltage results in a value greater than 1. A larger voltage is
required to give a realistic number.
3. A dramatic consequence of the uncertainty principle is that a particle confined in a
finite region of space cannot be exactly at rest! Find the minimum uncertainty in speed of
a particle of mass m that is confined to a one-dimensional box of width a. Use this to find
the minimum average kinetic energy of the particle. This is called the ground state energy
of a particle in a box. We will derive this minimum energy later from the Schrodinger
equation.
The position uncertainty of a particle in a box is the spread from the central point to
either side. So
a
Δx =
2
From the uncertainty principle,


ΔxΔp ≥ ⇒ Δp ≥
2
2Δx
p = mv ⇒ Δp = mΔv

⇒ mΔv ≥
2Δx


Δv ≥
=
2mΔx ma
The kinetic energy of a particle is
1
KE = mv 2
2
The average kinetic energy must be greater than
1
2
KEave ≥ mΔv 2 =
2
2ma 2
4. A particle is described by the wave function
−L
L
$
≤x≤
& 3A, for
ψ (x) = %
4
4
&' 0,
otherwise
a) Make a sketch of the probability density function, P(x) as a function of x.
b) Using the sketch, normalize the wave function and determine the constant A in terms
of L.
c) Use the sketch to determine the probability that the particle lies in the region x > L/3.
−L
L
$ 2
≤x≤
&9A , for
(a) P(x) = ψ (x) = %
4
4
&' 0,
otherwise
2
P(x
)
9A
2
-L/4
L/4
x
(b) Normalization: The integral of P(x) over all values of x must be equal to 1. This is
equivalent to the requirement that the area under the P(x) curve must be equal to 1. From
the sketch, the area under the P(x) curve is
L
2
Area = 9A 2 = 1 ⇒ A =
2
9L
(c ) The probability for the particle to lie in the region x>L/3 is 0 since the function P(x)
is zero in this region.
5. A particle is described by the wave function
2π x
−L
L
%
for
≤x≤
' A cos
ψ (x) = &
L
4
4
'(
0,
otherwise
a) Normalize the wave function and determine the constant A in terms of L.
b) What is the probability that the particle lies between x = 0 and x= L/8?
1
B
Hint: ∫ B cos 2 (α x)dx = Bx +
sin(2α x)
2
4α
−L
L
+ 2
2 # 2π x &
for
≤x≤
- A cos %
(
$ L '
a) P(x) = ψ (x) = ,
4
4
0,
otherwise
.
2
Normalization: The integral of P(x) over all values of x must be equal to 1.
L/4
∞
1 2
A2 L
4π x
% 2π x (
P(x)dx
=
A
cos
dx
=
A
x
+
sin
'&
*)
∫−∞
∫
L
2
8π
L
−L/4
2
⇒A=
L/4
2
= A2
−L/4
L
=1
4
4
L
(b) Probability that the particle lies between x = 0 and x= L/8:
L /8
L /8
∫ P(x)dx = ∫
0
0
1 2
A2 L
4π x
# 2π x &
A cos %
dx
=
A
x
+
sin
(
$ L '
2
8π
L
2
L /8
2
= A2
0
L
L
1 1
+ A2
= +
= 0.41 = 41%
16
8π 4 2π