More Problems

Problem 1: 

The cutoff wavelength for the photoelectric effect in tungsten metal is 274 nm.
(a)  What is the stopping potential for the photoelectron when the system is illuminated with 200 nm light?
(b)  What is the De Broglie wavelength of the photoelectron when it is ejected from the material under these conditions?

Solution:

• Concepts:
  The photoelectric effect
• Reasoning:
  The maximum kinetic energy of the ejected electrons is given by E = hf - Φ.  If the maximum kinetic energy is x eV, then the stopping potential is x V.
  For the cutoff or maximum wavelength λ_c we have Φ =  hc/λ_c. 
• Details of the calculation:
  (a)  Here Φ = (1240 eV nm)/(274 nm) = 4.53 eV.  E = (1240 eV nm)/(200 nm) - 4.53 eV = 1.67 eV.
  The stopping potential is 1.67 V.
  
  (b) The De Broglie wavelength is λ = h/p.  We need to decide if we must use the relativistically correct expression for p, or if the non-relativistic expression suffices.  Here we have a non-relativistic situation.
  T = 1.67 eV = p²/(2m_e).
  λ = h/(2m_e 1.67 eV)^½ = hc/(2m_ec² 1.67 eV)^½
  = (1240eV nm)/(3.35*0.51*10⁶ eV²)^½ = 0.95 nm.

Problem 2:

For NaCl crystals the ions of Na and Cl are on the cubic lattice shown in the figure.  If the density of NaCl is 2.16 g/cm³, what are the distances between nearest, next-nearest, and next-next-nearest neighboring atoms? 
(Na atomic mass is 22.99g/mole and Cl atomic mass is 35.45 g/mole, Avogardro's constant is N_A = 6.022x10²³mol^-1).

Solution:

• Concepts:
  Density, geometry
• Reasoning:
  The density of NaCl is 2.16 g/cm³.
  The average atomic mass of Na and Cl is (22.99 + 35.45)/2 = 29.22 g/mole.
  1 cm³ contains 2.16/29.22 = 0.0739 average Na/Cl moles or 0.0739*N_A = 4.396x10²² atoms^.
• Details of the calculation:
  The number of atoms along the edge of a cube with volume of 1 cm is (4.396*10²²)^1/3 = 3.529*10⁷.
  If the distance between atoms along the edge is d, then 3.529*10⁷ d = 0.01 m.
  The lattice spacing d = 0.283 nm.
  Nearest neighbors are at distance d = 0.283 nm.
  Next-nearest neighbor distance is diagonally over a square side so the distance is √2*d = 0.4 nm.
  Next-next-nearest neighbors are diagonally over the cube so the distance is √3*d = 0.49 nm.

Problem 3:

A laser beam with wavelength of 633 nm is incident on a dark fiber.  The first dark fringe is observed at an angle θ = 7^o away from the incident direction.  Find the diameter of the fiber.

Solution:

• Concepts:
  Single slit diffraction
• Reasoning:
  The dark fringes in the diffraction pattern of a single slit are found at angles θ for which w sinθ = mλ, where λ is the wavelength of the light and m is an integer, m = 1, 2, 3, ... .
  The intensity at the screen is proportional to the square of the electric field amplitude. 
  -If we block the slit completely with an opaque blocker, the electric field at a large distance is zero. 
  -If we remove mask with the slit, the electric field at a large distance is that of the non-diffracted beam.
  What if we remove the mask and only leave the blocker of width w?  Using Huygens' principle we have
  E_{mask with slit} + E_{blocker (no mask)} = E_{non-diffracted beam}.
  Here E_{mask with slit} is the field produced by sources at locations of the mask and E_{blocker (no mask)} is the field produced by source at locations of the blocker.
  Therefore E_{blocker (no mask)} = E_{non-diffracted beam} - E_{mask with slit}.
  For a laser beam the divergence angle θ₀ is small, and for angles θ > θ₀ we have
  E_{blocker (no mask)} = -E_{mask with slit}.
  For angles θ > θ₀ the average intensity, which is proportional to the square of the electric field, therefore is the same as that for the single slit.  Dark fringes in the diffraction pattern are found at angles θ for which w sinθ = mλ.
• Details of the calculation:
  Here w = 633 nm/sin(7^o) = 5194 nm.

Problem 4:

A 3-kg sphere dropped through air has a terminal speed of 25 m/s.  Assume that the drag force is proportional to the velocity.  The sphere is attached to a spring of force constant 400 N/m and starts to oscillate with an initial amplitude of 20 cm.
(a)  When will the amplitude be 10 cm?
(b)  How much energy will have been lost when the amplitude is 10 cm?

Solution:

• Concepts:
  The damped harmonic oscillator
• Reasoning:
  The equation of motion is F = -kx - cdx/dt.  Given the terminal speed we can solve mg = cv_terminal for c.
• Details of the calculation:
  (a)  When considering masses oscillating on vertical springs in a uniform gravitational field, the gravitational force does not appear in the equations of motions if we write these equations in terms of the displacements from the equilibrium positions in the gravitational field.
  The equation of motion for the mass is d²x/dt² = -ω₀²x - (c/m)dx/dt.
  Here ω₀² = k/m = 133.33 s^-2 and c = mg/v_terminal = 1.1772 kg/s.
  Let x = Aexp(iΩt).  (The real part is implied.)  Then
  -Ω² + 2k/m + iΩc/m = 0,  Ω = ic/(2m) ± (ω₀² - c²/(2m)²)^½.
  The position as a function of time is given by
  x = A exp(-ct/(2m))cos(ωt + φ), with ω = (ω₀² - c²/(2m)²)^½.
  ω₀ = 11.547/s  >> c/(2m) = 0.196/s, so the period of the oscillation is T ≈ 2π/ω₀ = 0.542 s is much smaller than the inverse of the damping coefficient 1/γ = 2m/c = 5.1 s.  We have under-damped motion.
  
  The initial amplitude is A = 20 cm at t = 0 so φ = 0.
  Solve exp(-ct/(2m)) = ½ for the time t when the amplitude is 10 cm.
  exp(-ct/(2m)) = ½,  ct/(2m) = ln(2),  t = 2mln(2)/c = 3.53 s.

  (b) The total energy of the mass is E = ½kA².  The initial energy is 8 J.
  When the amplitude has been reduced to ½ the initial amplitude, 3/4 of the initial energy or 6 J has been lost.