Physics 249 Midterm Exam 1 Solutions
Oct 12th 2012
1) An electron of kinetic energy 45keV moves in a circular orbit perpendicular to a magnetic field of
0.325T. Find the radius of the orbit in meters
(15 points)
𝐹𝑐 = 𝑚𝑎𝑐 = 𝑚
𝑣2
= 𝑞𝑣𝐵 = 𝐹𝐵
𝑟
There are two ways to numerically solve this. Both pay attention to the fact that keV is not a standard
metric units.
8
2
2𝐸
2𝐸𝑐 2
√2 ∗ 45𝑘𝑒𝑉 ∗ (3.00𝑥10 𝑚/𝑠)
𝑚𝑣 𝑚√ 𝑚
𝑚 √ 𝑚𝑐 2
1
511𝑘𝑒𝑉
𝑅=
=
=
=
= 2.2x10−3 𝑚
𝑞𝐵
𝑞𝐵
𝑞 𝐵
1.76x1011 C/kg
0.325𝑇
In this case I canceled out the keV units leaving only metric units.
𝐸𝑒
√2𝑚2 𝐸 𝑒
√2 𝐸 𝑒
√2 𝐸
√2 45𝐾𝑒𝑉
𝑚√2
𝑚𝑣
𝑝
√2𝑚𝐸
𝑒𝑚
𝑒𝑚
𝑒𝑚
𝑒
𝑒
𝑅=
=
=
=
=
= 𝑒
=
=
=
11
𝑞𝐵 𝑞𝐵
𝑞𝐵
𝑒𝐵
𝑒𝐵
𝑒
C/kg 0.325𝑇
√1.76x10
𝐵
√ 𝐵
𝑚
𝑚
= 2.2x10−3 𝑚
In this case I use the definition of electron volts. 1 eV is the energy of an electron accelerated though a
potential difference of 1V. The units are electron-charge*Volts. By dividing by the electron charge you
convert to Volts, which is a standard metric unit.
2) Derive the units of the fine structure constant.
(15 points)
𝛼=
𝑘𝑒 2
ℏ𝑐
There are various easy ways to do this.
ke2 is the top portion of the Electric force equation for a point charge, F= ke2/r2 (covered in Rutherford
scattering).
ke2 = Fr2 = [(force*distance)*distance] = [Energy*distance]
hc: energy*distance.
Therefore alpha is dimensionless.
A second method
The potential of an electric field (not potential energy) is:
𝑉=
𝑘𝑒
[𝑉𝑜𝑙𝑡𝑠]
𝑟
𝑒𝑉𝑟 = 𝑘𝑒 2 [𝑒𝑉 ∗ 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒] = 𝑘𝑒 2 [𝑒𝑛𝑒𝑟𝑔𝑦 ∗ 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒]
hc: energy*distance
Therefore alpha is dimensionless. (this methods puts the numerator and denominator in the same energy
units)
The fact that alpha is dimensionless is often cited as a reason why it a truly fundamental constant.
3) Demonstrate that the wave function of the lowest energy level of the particle in a box obeys the
uncertainty relation by calculating Δ𝑥 and Δp using expectation values.
For the potential use:
V(x) = 0, 0 < x < L
V(x) = infinity x < 0, x > L
(30 points)
The wave function must satisfy the Schrodinger equation inside the box and be zero at x=0 and x=L
The general formula for the wave function is:
2
𝜋
𝜓𝑛 (𝑥) = √ 𝑠𝑖𝑛 (𝑛 𝑥)
𝐿
𝐿
from x=0 to x=L
The normalization can be found from normalizing the n=1 ground state wave function.
(No point taken off is the student starts from this wave function.)
The lowest energy level should result in a minimum Δ𝑥 and Δp consistent with the uncertainty principle.
2
𝜋
𝜓1 (𝑥) = √ 𝑠𝑖𝑛 ( 𝑥)
𝐿
𝐿
The uncertainty principle is defined by multiplying the standard deviation of x and p. (chapter 5.5)
(The definition of the standard deviation from the mean was defined in the lecture on expectation values):
Note: 3 of the 4 integrals we to perform to evaluate these expectation values have trivial solutions. I will
evaluate 2 of 4 since the easy solution to the 3rd integral is not so obvious.
𝜎𝑥 = √⟨(𝑥 − ⟨𝑥⟩)2 ⟩ = √⟨𝑥 2 − 2𝑥⟨𝑥⟩ + ⟨𝑥⟩2 ⟩ = √⟨𝑥 2 ⟩ − ⟨𝑥⟩2
∞
𝐿
2
𝜋
2
𝜋
⟨𝑥 2 ⟩ = ∫ 𝜓 ∗ (𝑥, 𝑡)𝑥 2 𝜓(𝑥, 𝑡)𝑑𝑥 = ∫ √ 𝑠𝑖𝑛 ( 𝑥) 𝑥 2 √ 𝑠𝑖𝑛 ( 𝑥) 𝑑𝑥
𝐿
𝐿
𝐿
𝐿
−∞
0
⟨𝑥 2 ⟩ =
2 𝐿 2 2 𝜋
∫ 𝑥 𝑠𝑖𝑛 ( 𝑥) 𝑑𝑥
𝐿 0
𝐿
using
⟨𝑥 2 ⟩ =
2 𝑥3
𝐿𝑥 2
𝐿3
𝜋
𝑥𝐿2
𝜋
( −(
− 3 ) 𝑠𝑖𝑛 (2 𝑥) − 2 𝑐𝑜𝑠 (2 𝑥)) |𝐿0
𝐿 6
4𝜋 8𝜋
𝐿
4𝜋
𝐿
⟨𝑥 2 ⟩ =
2 𝐿3
𝐿3
1
1
( − 2 ) = 𝐿2 ( − 2 )
𝐿 6 4𝜋
3 2𝜋
The wave function is symmetric and centered at L/2 so:
𝐿 2 𝐿2
⟨𝑥⟩2 = ( ) =
2
4
1
1
2
2
2
⟨𝑥 ⟩ − ⟨𝑥⟩ = 𝐿 ( − 2 )
12 2𝜋
√⟨(𝑥 − ⟨𝑥⟩)2 ⟩ = 𝐿√(
1
1
−
)
12 2𝜋 2
𝜎𝑝 = √⟨(𝑝 − ⟨𝑝⟩)2 ⟩ = √⟨𝑝2 ⟩ − ⟨𝑝⟩2
∞
⟨𝑝2 ⟩ = ∫ 𝜓 ∗ (𝑥, 𝑡) (−ℏ2
−∞
𝜕2
) 𝜓(𝑥, 𝑡)𝑑𝑥
𝜕𝑥 2
𝐿
2
𝜋
𝜕2
2
𝜋
⟨𝑝2 ⟩ = ∫ √ 𝑠𝑖𝑛 ( 𝑥) (−ℏ2 2 ) √ 𝑠𝑖𝑛 ( 𝑥) 𝑑𝑥
𝐿
𝐿
𝜕𝑥
𝐿
𝐿
0
⟨𝑝2 ⟩ =
2𝜋𝜋 2 𝐿
𝜋
ℏ ∫ 𝑠𝑖𝑛2 ( 𝑥) 𝑑𝑥
𝐿𝐿𝐿
𝐿
0
using
⟨𝑝2 ⟩ =
2𝜋 2 ℏ2 𝑥 𝐿 𝜋 2 ℏ2
( ) |0 = 2
𝐿3
2
𝐿
Note you also get ⟨𝑝2 ⟩ from first energy level. Since energy is a constant proportional to 𝑝2 and ⟨𝑝2 ⟩ =
⟨2𝑚𝐸⟩ = 2𝑚𝐸.
The momentum has equal probability to be positive or negative so on average is zero:
⟨𝑝⟩2 = 0
⟨𝑝2 ⟩ − ⟨𝑝⟩2 =
𝜋 2 ℏ2
𝐿2
√⟨(𝑝 − ⟨𝑝⟩)2 ⟩ =
Δ𝑥Δ𝑝 = 𝐿√(
𝜋ℏ
𝐿
1
1 𝜋ℏ
1
−
)
= 0.568ℏ > ℏ
12 2𝜋 2 𝐿
2
Which obeys the uncertainty relationship and matches the expectation that the lowest energy level will have
the smallest uncertainty since the uncertainty in momentum is the smallest.
4) Explain why the infinite potential well wave function satisfied the uncertainty principle, but with slightly
larger uncertainty in x and p, while the lowest energy state of the harmonic oscillator satisfies the
uncertainty principle with the minimum possible uncertainty in x and p.
(15 points)
The lowest energy state of the harmonic oscillator has a Gaussian probability distribution in space, which is
the same probability distribution for the shape of wave packet. For the wave packet (or quantum harmonic
oscillator) when you calculate the distributions of momentums you also find a Gaussian distribution.
As explained in the text book the combination of Gaussian distributions in both space and momentum for
the wave packet results in the minimum possible uncertainty. Any other probability distribution in space
will have a larger uncertainty. The first cos function solution is similar to a Gaussian in that it is peaked in
the center of the potential and falls off to either side to but not exactly the same resulting in a slightly larger
uncertainty.
Another argument that can be made is that any system that highly localizes x or p, in this case x is highly
constrained to be inside the box, tends to make the other value, p or x respectively, have higher uncertainty
driving the uncertainty over the minimum value.
5) For an electron in a finite potential well with L=0.001nm and V=10MeV how many bound state
solutions are there. The conditions for the quantized energy solutions are:
𝑘𝑛 𝐿
𝛼
tan (
)=
2
𝑘𝑛
and
𝑘𝑛 𝐿
𝛼
−cot (
)=
2
𝑘𝑛
(25 points)
recast this as
𝑘𝑛 𝐿
𝑘𝑛 𝐿
𝑚𝐿2 𝑉0
𝑘𝑛 𝐿 2
tan (
)=√
−
(
)
2
2
2ℏ2
2
and
−
𝑘𝑛 𝐿
𝑘𝑛 𝐿
𝑚𝐿2 𝑉0
𝑘𝑛 𝐿 2
cot (
)=√
−
(
)
2
2
2ℏ2
2
together tan and cot will have zeros at
𝑘𝑛 𝐿
𝜋
=𝑛
2
2
including n=0 where the 1st tangent solution starts.
and any tan or cot line starting there will cross the parabolic curve if the maximum of the parabolic curve is
larger.
the maximum solution is at
𝑘𝑛 𝐿
𝑚𝐿2 𝑉0
4𝜋 2 𝑚𝑐 2 𝐿2 𝑉0
2𝜋 2 (0.511𝑀𝑒𝑉)(1000𝑓𝑚)2 10𝑀𝑒𝑉
𝜋
=√
=√
=√
= 8.099 = 5.16
2
2
2
2
(1240𝑀𝑒𝑉𝑓𝑚)
2
2ℏ
2ℎ 𝑐
2
6 bound states exist.