Problem#1
A beam of neutrons that all have the same energy scatters from atoms that have a spacing of 0.0910 nm in the surface plane of a crystal. The m = 1 intensity maximum occurs when the angle θ in Fig. 1 is 28.60. What is the kinetic energy (in electron volts) of each neutron in the beam?
Answer:
The intensity maxima are located by Eq. (39.4). Use λ = h/p for the wavelength of the neutrons. For a particle, p = √(2mE).
For neutron, m = 1.67 x 10-27 kg
For m = 1, λ = d sinθ = h/√(2mE)
E = h2/(2md2sin2θ)
E = (6.63 x 10-34 J.s)2/[2(1.675 x 10-27 kg)(9.10 x 10-11 m)2sin2(28.60)] = 6.91 x 10-20 J = 0.432 eV
The neutrons have λ = 0.0436 nm, comparable to the atomic spacing.
Problem#2
A beam of 188-eV electrons is directed at normal incidence onto a crystal surface as shown in Fig. 2. The m = 2 intensity maximum occurs at an angle θ = 60.60. (a) What is the spacing between adjacent atoms on the surface? (b) At what other angle or angles is there an intensity maximum? (c) For what electron energy (in electron volts) would the m = 1 intensity maximum occur at θ = 60.60 For m = 2 this energy, is there an intensity maximum? Explain.
Answer:
Intensity maxima occur when d sinθ = mλ, λ = h/p = h/√(2ME) so dsinθ = mh/√(2ME)
Here m is the order of the maxima, whereas M is the mass of the incoming particle.
(a) dsinθ = mh/√(2ME)
dsin60.60 = (2)(6.63 x 10-34 J.s)/√[2(9.11 x 10-31 kg)(188 eV)(1.60 x 10-19 J/eV)]
d = 2.06 x 10-10 m = 0.206 nm.
(b) m = 1 also gives a maxima.
(2.06 x 10-10 m)sinθ = (1)(6.63 x 10-34 J.s)/√[2(9.11 x 10-31 kg)(188 eV)(1.60 x 10-19 J/eV)]
sin θ = 0.4352
θ = 25.80.
This is the only other one. If we let 3, m ≥ then there are no more maxima.
(c) E = m2h2/(2Md2sin2θ)
E = (1)2(6.63 x 10-34 J.s)2/[2(9.11 x 10-31 kg) (2.06 x 10-10 m)2 sin2(60.60)]
E = 7.49 x 10-18 J = 46.8 eV
Using this energy, if we let m = 2, then sinθ > 1. Thus, there is no m = 2 maximum in this case.
As the energy of the electrons is lowered their wavelength increases and a given intensity
maximum occurs at a larger angle.
Problem#3
A CD-ROM is used instead of a crystal in an electrondiffraction experiment. The surface of the CD-ROM has tracks of tiny pits with a uniform spacing of 1.60µm. (a) If the speed of the electrons is 1.26 x 104 m/s at which values of θ will the m = 1 and m = 2 intensity maxima appear? (b) The scattered electrons in these maxima strike at normal incidence a piece of photographic film that is 50.0 cm from the CD-ROM. What is the spacing on the film between these maxima?
Answer:
The condition for a maximum is d sinθ = mλ.
λ = h/p = h/Mv
So, θ = arcsin(mh/dMv)
Here m is the order of the maximum, whereas M is the incoming particle mass.
(a) m = 1 → θ1 = arcsin(h/dMv)
θ1 = arcsin{(6.63 x 10-34 J.s)/[(1.60 x 10-9 m)(9.11 x 10-31 kg)(1.26 x 104 m/s)]} = 2.070
for m = 2,
θ2 = arcsin{(2)(6.63 x 10-34 J.s)/[(1.60 x 10-9 m)(9.11 x 10-31 kg)(1.26 x 104 m/s)]} = 4.140
(b) For small angles (in radians!) y ≈ Dθ, so y1 ≈ (50 cm)(2.070)(π radians/1800) = 1.81 cm
y2 ≈ (50 cm)(4.140)(π radians/1800) = 3.61 cm
and
y2 – y1 = 3.61 cm – 1.81 cm = 1.80 cm
Problem#4
(a) In an electron microscope, what accelerating voltage is needed to produce electrons with wavelength 0.0600 nm? (b) If protons are used instead of electrons, what accelerating voltage is needed to produce protons with wavelength 0.0600 nm? (Hint: In each case the initial kinetic energy is negligible.)
Answer:
(a) We use
eV = K = p2/2m = (h/λ)2/2m
so, V = (h/λ)2/2me
V = (6.63 x 10-34 J.s)2/[2(9.11 x 10-31 kg)(1.6 x 10-19 C)(0.060 x 10-9 m)2] = 419 V
(b) the voltage is reduced by the ratio of the particle masses,
(419 V)(9.11 x 10-31 kg)/(1.67 x 10-27 kg) = 0.229 V
Problem#5
You want to study a biological specimen by means of a wavelength of 10.0 nm, and you have a choice of using electromagnetic waves or an electron microscope. (a) Calculate the ratio of the energy of a 10.0-nm-wavelength photon to the kinetic energy of a 10.0-nm-wavelength electron. (b) In view of your answer to part (a), which would be less damaging to the specimen you are studying: photons or electrons?
Answer:
For a photon Eph = hc/λ = 1.99 x 10-25 J.m/λ.
For electron Ee = p2/2m = (h/λ)2/2m = h2/2mλ2
(a) photon, Eph = (1.99 x 10-25 J.m)/(10.0 x 10-9 m) = 1.99 x 10-17 J
Electron, Ee = (6.63 x 10-34 J.s)2/[2(9.11 x 10-31 kg)(10.0 x 10-9 m)2] = 2.41 x 10-21 J
Eph/Ee = (1.99 x 10-17 J)/(2.41 X 10-21 J) = 8.26 x 103
(b) The electron has much less energy so would be less damaging.
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