Problem 1:
Consider an atomic nucleus.
(a) Why do nuclei tend to have approximately the same number of neutrons and
protons?
(b) As you increase element atomic number, why does the ratio of neutrons to
protons increase?
(c) Why do light nuclei release energy via fusion while heavy nuclei release
energy via fission?
Solution:
- Concepts:
Nuclear properties
- Reasoning:
(a) Nucleons are fermions and obey the Pauli exclusion principle. The shell
model, with separate energy shell for protons and neutrons explains that adding
too many neutrons forces them to occupy higher energy levels and therefore
lowers the binding energy and leads to beta decay. A nucleus consisting of only
neutrons cannot exist.
(b) The dominant interactions between nucleons in a
nucleus are the short-range nuclear force and the long-range electrostatic
force. Nucleons are only attracted by other nucleons in their neighborhood, but
protons are repelled by all other protons in the nucleus. As nuclei with more
than ~56 nucleons grow larger, the net attractions between nucleons stays
roughly the same, while the net repulsion increases significantly. Decreasing
the number of protons decreases the repulsion and results in a nucleus with a
larger binding energy.
(c) When a large nucleus fissions into roughly two equal parts, the net
attractions between nucleons stays roughly the same, while the net repulsion
decreases significantly. Electrostatic potential energy is released.
When two light nuclei fuse, the net strong attractions between nucleons and the
net repulsion between the protons increases, but the nuclear force is the much
stronger force. Nuclear potential energy is released.
Problem 2:
For this one-dimensional problem
let the x-axis point downward. A vertical spring has a spring constant
k = 48 N/m. At t = 0 a force F(t) = (51 N)sin(4t) (t ≥ 0 with t in seconds) is
applied to a 30 N weight which hangs in equilibrium at the end of the spring.
Neglecting damping, find the position of the weight at any time t.
Solution:
- Concepts:
Driven oscillations
- Reasoning:
We have motion in one dimension.
The total force on the mass is F = -kx + mg + F0sinωt = md2x/dt2.
Here x is the displacement of the end of the spring from its unstretched
position, when no mass is attached.
- Details of the calculation:
d2x/dt2 + (k/m)(x - mg/k) = (F0/m)sinωt.
Let x' be the displacement from the equilibrium position of the mass, x' = x -
mg/k.
d2x'/dt2 + (k/m)x' = (F0/m)sinωt.
A particular solution of the inhomogeneous equation is
x' = A sinωt. A = (F0/m)/(ω02
- ω2), with ω02 = k/m.
F0/m = (51 N)*(9.8 m/s2)/(30 N) = 16.66 m/s2.
ω02 = 15.68/s2. ω0 = 3.95/s.
ω02
- ω2 = (48*9.8/30 - 16) s-2 =
-0.32/s2.
A = -52 m. x' = -52 m sin(4t).
The driving frequency is just above the resonance frequency.
To find the most general solution we add the homogeneous solution x' = Csin(ω0t
+ δ).
Most general solution: x(t) = Csin(ω0t + δ)
- (52 m)sin(ωt).
Initial conditions: x' = 0
and dx'/dt = 0 at t = 0.
Csin(δ) = 0 --> δ = 0.
ω0C- ω(52 m)
= 0 --> C = (ω/ω0)(52
m).
x'(t) = (ω/ω0)(52
m)sin(ω0t) - (52 m)sin(ωt),
dx'/dt = ω(52
m)cos(ω0t) - ω (52 m)cos(ωt)
satisfy the initial conditions.
x'(t) =
(1.01)(52
m)sin(3.95t) -
(52m)sin(4t)
is the displacement of the mass from its equilibrium position as a function of
time. A graph of x'(t) vs t will show "beats".
Problem 3:
In 2017 ETH researchers succeeded in shortening the pulse duration of an
X‑ray laser to only 43 attoseconds to produce the world's shortest laser pulse.
The soft X-ray laser pulse has a very large spectral bandwidth. What is
the uncertainty in the momentum of a photon in this pulse?
Solution:
- Concepts:
The uncertainty principle
- Reasoning:
For any wave, ΔωΔt ≈ 1, ΔkΔx ≈ 1.
- Details of the calculation:
Here Δt = 43*10-18 s is the maximum observation time. Δω = 1/Δt. ћΔω/c
= Δp = ћ/(cΔt).
Δp = (1.05*10-34 Js)/(3*108 m/s * 4.3*10-18 s)
= 8.14*10-26 kgm/s = 153 eV/c.
or
Δx = cΔt = maximum length of pulse. Δk = 1/Δx = 1/(cΔt). ћΔk = Δp =
ћ/(cΔt).
Problem 4:
Assume white light (λ = 400 nm - 650 nm) is incident at an angle of 60o
with respect to the normal from air
on a thin film of oil on water. The thickness of the oil film is t = 100
nm, and the indices of refraction of water and oil are nwater = 1.33
and noil = 1.5, respectively. Which wavelength(s) will dominate
in the reflected light?
Solution:
- Concepts:
Thin film interference
- Reasoning:
When a light wave reflects from a medium with a larger index of refraction,
then the phase shift of the reflected wave with respect to the incident wave is
π or 180o.
When a light wave reflects from a medium with a smaller index
of refraction, then the phase shift of the reflected wave with respect
to the incident wave is zero.
- Details of the calculation:
Waves incident at an angle θi on the air oil interface are refracted
as they enter the oil. The angle
of refraction θt is found from
Snell's law, nairsinθi = noilsinθt. If the
waves are reflected off the second interface, then they travel a
distance 2t/cosθt in the oil.
When they emerge again from the oil into the air and propagate parallel
to the waves reflected at the air-oil interface, then the total optical
path length difference is
2noil t/cosθt - 2t tanθtsinθi
= 2noil t/cosθt - 2t tanθt(noil/nair)sinθt
= (2noilt/cosθt)(1 - sin2θt)
= 2noil t cosθt = 2 noil t(1 - (nair/noil)2sin2θi)½.
(optical path length in a medium = n* actual path length in the medium)
We get a phase shift of π only for the wave reflected from the first interface.
For constructive interference we therefore need 2noil t cosθt
= (m + ½)λ, m = 0, 1, 2, ... .
Using the given θi, t, and noil, we find (m + ½)λ = 245
nm.
The only wavelength in the visible region of the spectrum for which we will have constructive interference at 60o
reflection is λ = 490 nm.
Problem 5:
The radioactive isotope 229Th is an
α emitter with a half-life t½ Th
= 7,300 y. Its daughter, 225Ra, is a β emitter
with a half-life t½ Ra = 14.8 d, which is much
shorter than the parent's.
(a) Show that the number of 225Ra nuclei obey the differential
equation
dNRa/dt = NTh/τTh
- NRa/τRa,
where NTh is the number of 229Th nuclei and NRa is the number
of 225Ra nuclei, and τTh
and τRa are their mean lives, respectively.
(b) Suppose that at time t = 0 there are NTh0 nuclei
of the parent isotope 229Th and no nuclei of the daughter isotope
225Ra. Find NRa for t > 0.
(c) Show that after several years, NRa = NTh
τRa/τTh.
Solution:
- Concepts:
Radioactive decay
- Reasoning:
The decay of an unstable nucleus is a random (probabilistic) process. The rate
at which nuclei decay is proportional to the number N of nuclei present.
- Details of the calculation:
(a) dNTh/dt = -NTh/τTh, NTh(t) = N0Thexp(-t/τTh).
dNRa/dt = NTh/τTh - NRa/τRa.
(feed rate plus decay rate)
dNRa/dt = (1/τTh)N0Thexp(-t/τTh) - NRa/τRa.
(b)
Try a solution NRa = C exp(-t/τTh) + D exp(-t/τRa).
D = -C, since NRa(0) = 0.
NRa = C(exp(-t/τTh) - exp(-t/τRa)),
dNRa/dt = -(C/τTh)exp(-t/τTh) + (C/τRa)exp(-t/τRa))
But we also have
dNRa/dt
= (1/τTh)N0Thexp(-t/τTh) - C(exp(-t/τTh) - exp(-t/τRa))/τRa.
Equating we get
-C/τTh = N0Th/τTh - C/τRa,
C = τRaN0Th/(τTh- τRa).
NRa(t) = [τRaN0Th/(τTh - τRa)] (exp(-t/τTh)
- exp(-t/τRa)).
(c) τTh
>> τRa, τTh
>> several years, τRa << several years.
τRaN0Th/(τTh - τRa) ≈ τRaN0Th/τTh, exp(-t/τTh)
≈ 1, exp(-t/τRa) ≈ 0.
NRa(t) ≈ (τRa/τTh)N0Thexp(-t/τTh
) = (τRa/τTh)NTh.