3 February 2000 Dr. R. Egerton

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electron charge = 1.60 ´ 10^-19 C electron rest mass = 9.11 ´ 10^-31 kg

electron rest energy = 0.511 MeV speed of light in vacuum = 3.00 ´ 10⁸ m/s

Rydberg energy R = 13.6 eV Planck’s constant h = 6.63 x 10^-34 J s

1. An electron has a kinetic energy of 100 eV.

(a) Calculate its de Broglie wavelength.

(b) Calculate the smallest angle of diffraction (measured relative to the incident direction) when this electron interacts with a layer of gold atoms of spacing 0.41 nm.

(c) In addition to their spacing, what information about the gold atoms could be obtained from this kind of diffraction experiment ?

2. A photon (wavelength = 10^-12 m) is elastically scattered through an angle of 60 deg. by a single particle.

(a) Calculate the transverse component of momentum (perpendicular to the incident direction) transferred to the particle.

(b) Estimate the image resolution possible by focussing many photons, whose scattering angles vary from zero up to ± 60 degrees.

3. Photons (wavelength 600 nm) pass through two parallel slits (D = 3 mm apart) and are viewed on a screen (placed a distance L = 1 m from the slits).

1. Sketch a graph showing photon intensity as a function of distance x from the center of the pattern formed on the screen.
2. Estimate the spacing of the maxima in your graph.
3. How does this graph change as the photon intensity is reduced towards zero ?
4. How are the results of this experiment explained according to the Copenhagen interpretation of quantum mechanics ?

4. Specify precisely (by means of a diagram or in words) the electrostatic potential V(x) which gives rise to each of the following electron wavefunctions.

1. y = A exp(-a x²)
2. y = A sin(2p x/l ) for all x

(c) y = A sin(2p x/L) for 0<x<L, y = 0 for other values of x

(d) Write down the quantum number n for case (a), and also for case (c).

(e) For case (c) above, calculate the energy (in eV) of the photon emitted when the electron makes a transition to its lowest-energy state, taking L = 0.8 nm

Ph 312 Mid-Term (Feb. 2000) Answers (R. Egerton)

1. (a) K= (100)(1.6E-19) J = 0.5 m v^2

v^2 = (2)(1.6E-17)/(9.11E-31) = 3.51E13, v = 5.93E6 m/s

lambda = h/(mv) = (6.63E-34)/[(9.11E-31)(5.93E6)] = 1.23E-10 m

(b)For constructive interference, d sin(phi) = n (lambda)

Smallest phi corresponds to n=1,

giving sin(phi) = 1.23E-10/0.41E-9 = 0.30 , phi = 17.5 deg = 0.30 rad

(c) The experiment gives information about the symmetry of arrangement of the atoms.

2. (a) photon momentum p = h/lambda = 6.63E-22 kg m/s

transverse momentum transfer = p sin(theta) = (6.63E-22)(0.866) =5.74E-22 kg m/s

(b) spatial resolution = Dx ~ h / Dp = h/ [2p sin(theta)] ~ 6E-13 m

3. (a) The graph should show cosine-like oscillations, the central maximum at x = 0 and amplitude gradually decreasing as x becomes more positive or more negative.

(b) The maxima are given by n(lambda) = D sin(theta) ~ D (theta),

where theta ~ tan (theta) = x/L , giving x = n (L/D) (lambda) = n (1000/3)(600E-9)

Therefore spacing = (1000/3)(600E-9) = 2E-4 m = 0.2 mm

(c) The continuous oscillation of intensity would be replaced by discrete points, each representing arrival of one photon. The distribution of points would follow the original oscillations but with some statistical scatter.

(d) The Copenhagen interpretation says that the oscillation of intensity is an interference pattern arising from the wavelike nature of the photons within the apparatus. However, the discrete points on the graph indicate particle-like properties, which arise when the wave function "collapses" at the screen.

4. (a) The potential energy U of the electron is a parabola with its minimum at x = 0.

Therefore then electrostatic potential V (= U/-e) is a parabola with a maximum at x = 0.

(b) Sinusoidal oscillation for all x indicates a free electron: V = 0 (or a constant value).

(c)The potential energy is zero within the range 0<x<L and increases abruptly to infinity at x=0 and x=L. The potential V is zero but decreases to –(infinity) at x=0 and x=L.

(d) Case (a) corresponds to n = 0, according to the definition E = (n = ½) (hbar)(omega)

Case (c) to n = 2 (since a whole period of oscillation occurs over the length L).

The n=2 level has an energy E2 = (n^2)[(pi hbar/L)^2] / 2m

Therefore photon energy = E2 – E1 = (3)(9.4E-20) J = 1.76 eV