Problem 1:

Protons with kinetic energy of 1 GeV in a narrow beam are diffracted by Oxygen nuclei.  The first minimum is observed at an angle of 8.5 degree.  Calculate the radius of the Oxygen nuclei.

Solution:

• Concepts:
  Diffraction
• Reasoning:
  The intensity pattern of a circular aperture illuminated by a plane wave is called the "Airy Disk". 
  The first minimum occurs at an angle sinθ = 1.22 λ/d, where d is the diameter of the aperture.
  The intensity at a detector is proportional to the square of the wave amplitude. 
  -If we block the aperture completely with an opaque blocker, the amplitude at a large distance is zero. 
  -If we remove mask with the aperture, the amplitude at a large distance is that of the non-diffracted beam.
  If we remove the mask and only leave the blocker of diameter d, then we have
  E_{mask with hole} + E_{blocker (no mask)} = E_{non-diffracted beam}.
  Therefore E_{blocker (no mask)} = E_{non-diffracted beam} - E_{mask with hole}.
  For a narrow beam the divergence angle θ₀ is small, and for angles θ > θ₀ we have
  E_{blocker (no mask)} = -E_{mask with hole}.
  For angles θ > θ₀ the average intensity, which is proportional to the square of the amplitude, therefore is the same as that for the aperture.  The first minimum in the diffraction pattern is found at an angles θ for which sinθ = 1.22 λ/d.
• Details of the calculation:
• For protons with kinetic energy of 1 GeV we have  λ = h/p.
  E = m_pc² + T.  E² = m_p²c⁴ + 2Tm_pc² + T².
  E² = p²c² + m_p²c⁴,  p²c² = 2Tm_pc² + T².
  λ = hc/(2Tm_pc² + T²)^½
  = (1.24*10³ eV nm)/((10¹⁸ + 2*10⁹*0.938*10⁹)^½ eV) =  7.3*10^-7 nm.
  d = 1.22 λ/sinθ = 6*10^-6 nm = 6 fm
  The radius of the Oxygen nuclei is 3 fm.

Problem 2:

The Doppler effect is the change in the frequency of a wave caused by a moving source or observer.  Let's assume a source produces sound wave with the speed v and frequency f₀.
(a)  When the source moves toward a stationary observer with speed v_s, what is the frequency of the sound heard by the observer?
(b)  When the observer moves toward the stationary source with speed  v_o, what is the frequency of the sound heard by observer?
(c)  For a light wave with frequency f and speed c, show that the classical Doppler effect violates special relativity.  Use special relativity to derive the frequency of the light wave when the source is moving toward observer with the speed  v_s.

Solution:

• Concepts:
  The Doppler effect
• Reasoning:
  f = f₀(v - v_o)/(v - v_s),  f = observed frequency, f₀ = frequency of source,
  v = wave velocity,  v_o = velocity of observer, v_s = velocity of source.
  v_o and v_s are not the speeds, but the components of the observer's and the source's velocity in the direction of the velocity of the sound reaching the observer.
• Details of the calculation:
  (a)  f = f₀v/(v - |v_s|).
  (b)  f = f₀(v + |v_o|)/v.
  (c)  The classical Doppler effect assumes that wave moves in a medium, i.e. a special reference frame with speed v.  The speed of the wave is different in other reference frames that move themselves with respect to the medium.  That contradicts  a postulate of relativity.
  "In vacuum, light propagates with respect to any inertial frame and in all directions with the universal speed c."
  There are various ways to derive the Doppler formula for light waves. 
  For example:
  In reference frame K let orient the coordinate system so that the photon with frequency f moves in the xy-plane, its trajectory making an angle θ with the x-axis.
  Let reference frame K' move with velocity v/c = β in the positive x direction with respect to K.
  Then the Lorentz transformation of the momentum 4-vector of the photon from frame K to frame K' can be written as
  .
  We find hf'/c = γ(hf/c) - γβ(hf/c)cosθ,  f' = γf(1 - βcosθ).
  For the source moving toward observer with the speed θ = π, β = v_s/c, f' = γf(1 + v_s/c)).
  γ = 1/[(1 - v_s/c)(1 + v_s/c)] ½,  f' = f[(1 + v_s/c)/1 - v_s/c)]^½.

Problem 3:

In a deuterium-tritium (D-T) nuclear reaction 17.6 MeV of energy is released.  You are given a laser-fusion fuel pellet of mass 5 mg which is composed of equal parts (by mass) of deuterium and tritium.
(a)  If half the deuterium and an equal number of tritium nuclei participate in D-T fusion, how much total energy (in Joules) is released?
(b)  At what rate must pellets be fueled in a power plant with 3 GW thermal power output?
(c)  What amount of fuel would be needed to run the plant for 1 year?  Compare with the 3.6 megatons of coal needed to fuel a comparable coal-burning power plant.

Solution:

• Concepts:
  Mass of simple nuclei, ratios
• Reasoning:
  This problem just requires simple reasoning.
• Details of the calculation:
  (a)  Mass of tritium:  3 au;  # of tritium nuclei:  m_pellet/(6 au).
  Mass of deuterium:  2 au;  # of deuterium nuclei:  m_pellet/(4 au).
  # of fusions per pellet:   m_pellet/(8 au) = (5*10^-6 kg)/(8*1.67*10^-27 kg) = 3.74*10²⁰.
  Energy released per pellet:  E_pellet = 17.6 MeV*m_pellet/(8 au) = 6.58*10¹¹ MeV = 1*10⁹ J.
  (b)  To get 3*10⁹ J/s, we need (3*10⁹ J/s)/E_pellet = 3 pellets/s
  (c)  We need (3 pellets/s)(1 year)*( 5 mg) = 5.5 g.

Problem 4:

The potential energy of a mass m as a function of position x is given by U(x) = ax² + bx + c, with a, b, c positive constants.

(a)  Find the equilibrium position of the mass.
(b)  If the mass is released from rest at x = 0 at t = 0, find its position as a function of time assuming no resistive force is present.  What is the maximum kinetic energy of the mass?
(c)  If the mass is subject to a drag force F_d = -γv, (with γ² << 8am)  and is released from rest at x = 0 at t = 0, find the time when the amplitude of the motion is ¼ of the initial amplitude.

Solution:

• Concepts:
  The damped harmonic oscillator
• Reasoning:
  The potential energy function is that of a harmonic oscillator.
• Details of the calculation:
  (a)  F = -dU/dx = 2ax + b = 0.  x = -b/(2a).
  Let x' = x + b/(2a).  Then U = ax'² + constant, F = -2ax'.
  
  (b)  The mass moves in a harmonic oscillator potential.  The force obeys Hooke's law, F = -kx' with k = 2a.
  x'(0) = b/(2a),  v(0) = dx'/dt|₀ = 0,  x(t) =  [b/(2a)]cos(ω₀t),  ω₀ = (2a/m)^½.
  The amplitude of the motion is A =  b/(2a).
  The maximum kinetic energy of the mass is E = ½kA² = a[b/(2a)]² = b²/(4a).
  
  (c)  The damped harmonic oscillator.
  d²x'/dt² = -ω₀²x' - (γ/m)dx'/dt.
  Let x' = Aexp(iΩt).  (The real part is implied.)  Then
  -Ω² + 2a/m + iΩγ/m = 0,  Ω = iγ/(2m) ± (ω₀² - γ²/(2m)²)^½.
  x =  [b/(2a)]exp(-γt/(2m))cos(ωt), with ω = (ω₀² - γ²/(2m)²)^½.
  We have under-damped motion.
  exp(-γt/(2m)) = ¼,  γt/(2m) = ln(4),  t = 2mln(4)/γ.