Problem#1
Find the wavelengths of a photon and an electron that have the same energy of 25 eV. (Note: The energy of the electron is its kinetic energy.)
Answer:
Both for particles with mass (electrons) and for massless particles (photons) the wavelength is
related to the momentum p by λ = h/p. But for each type of particle there is a different expression that relates the energy E and momentum p.
For an electron E = ½ mv2 = p2/2m but for a photon E = pc.
Photon p = E/c and p = h/λ so h/λ = E/c and
λ = hc/E = (1.24 x 10-6 eV.m)/(25 eV) = 49.6 nm
Electron: Solving for p gives p = √(2mE). This gives
p = √{2(9.11 x 10-31 kg)(25 eV)(1.6 x 10-19 J/eV)} = 2.70 x 10-24 kg.m/s
The wavelength is therefore
λ = h/p = (6.63 x 10-34 J.s)/(2.70 x 10-24 kg.m/s) = 0.245 nm
Problem#2
(a) What accelerating potential is needed to produce electrons of wavelength 5.00 nm? (b) What would be the energy of photons having the same wavelength as these electrons? (c) What would be the wavelength of photons having the same energy as the electrons in part (a)?
Answer:
The acceleration gives momentum to the electrons. We can use this momentum to calculate
their de Broglie wavelength.
The kinetic energy K of the electron is related to the accelerating voltage V by K = eV. For an electron E = ½ mv2 = p2/2m and λ = h/p. for a photon E = hc/λ.
(a) For an electron p = h/λ = (6.63 x 10-34 J.s)/(5.00 x 10-9 m) = 1.33 x 10-25 kg.m/s and
E = p2/2m = (1.33 x 10-25 kg.m/s)2/{2(9.11 x 10-31 kg)} = 9.71 x 10-21 J.
V = K/e = (9.71 x 10-21 J)/(1.60 x 10-19 C) = 0.0607 V
The electrons would have konetic energy 0.0607 Ev.
(b) E = hc/λ = (1.24 x 10-6 eV.m)/(5.00 x 10-9 m) = 248 eV
(c) λ = hc/E = (6.63 x 10-34 J.s)(3.00 x 108 m/s)/(9.71 x 10-21 J) = 20.5 µm
Problem#3
Through what potential difference must electrons be accelerated so they will have (a) the same wavelength as an x ray of wavelength 0.150 nm and (b) the same energy as the x ray in part (a)?
Answer:
λ = h/p. Apply conservation of energy to relate the potential difference to the speed of the
electrons.
The mass of an electron is m = 9.11 x 10-31 kg. The kinetic energy of a photon is E = hc/λ.
(a) λ = hc/E → v = h/mλ. Energy conservation: e∆V = ½ mv2
∆V = mv2/2e = m(h/mλ)2/2e = h2/2emλ2
∆V = (6.63 x 10-34 J.s)2/[2(1.6 x 10-19 C)(9.11 x 10-31 kg)(0.15 x 10-9 m)2] = 66.9 V
(b) Ephoton = hf = hc/λ = (6.63 x 10-34 J.s)(3.00 x 108 m/s)/(0.15 x 10-9 m) = 1.33 x 10-15 J
e∆V = K = Ephoton
∆V = Kphoton/e = 1.33 x 10-15 J/1.6 x 10-19 C = 8310 V
Problem#4
(a) Approximately how fast should an electron move so it has a wavelength that makes it useful to measure the distance between adjacent atoms in typical crystals (about 0.10 nm)? (b) What
is the kinetic energy of the electron in part (a)? (c) What would be the energy of a photon of the same wavelength as the electron in part (b)? (d) Which would make a more effective probe of smallscale structures: electrons or photons? Why?
Answer:
For electron, λ = h/p and K = ½ mv2. For a photon, E = hc/λ. The wavelength should be 0.10 nm.
(a) λ = 0.10 nm. p = mv = h/λ so v = h/(mλ) = 7.3 x 106 m/s.
(b) K = ½ mv2 = ½ (9.11 x 10-31 kg)(7.3 x 106 m/s)2 = 2.427 x 10-17 J = 152 eV
(c) E = hc/λ = (6.63 x 10-34 J.s)(3.00 x 108 m/s)/(0.10 x 10-9 m) = 1.989 x 10-15 J = 12341 eV = 12 keV
(d) The electron is a better probe because for the same λ it has less energy and is less damaging to the structure being probed.
Problem#5
A beam of electrons is accelerated from rest through a potential difference of and then passes through a thin slit. The diffracted beam shows its first diffraction minima at ±11.50 from the original direction of the beam when viewed far from the slit. (a) Do we need to use relativity formulas? How do you know? (b) How wide is the slit?
Answer:
The electrons behave like waves and are diffracted by the slit.
We use conservation of energy to find the speed of the electrons, and then use this speed to find
their de Broglie wavelength, which is λ = h/mv. Finally we know that the first dark fringe for single-slit diffraction occurs when a sin θ = λ.
(a) Use energy conservation to find the speed of the electron:
½ mv2 = eV
½ (9.11 x 10-31 kg)v2 = (1.6 x 10-19 C)(100 V)
v2 = 3.513 x 1013 m2/s2
v = 5.93 x 106 m/s
which is about 2% the speed of light, so we can ignore relativity.
(b) First find the de Broglie wavelength:
λ = h/mv = (6.63 x 10-34 J.s)/[(9.11 x 10-31 kg)(5.93 x 106 m/s)] = 1.23 x 10-10 m = 0.123 nm
For the first single-slit dark fringe, we have a sin θ = λ, which gives
a = λ/sinθ = (1.23 x 10-10 m)/sin 11.50 = 6.16 x 10-10 m = 0.616 nm
The slit width is around 5 times the de Broglie wavelength of the electron, and both are much
smaller than the wavelength of visible light.
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