Problem#1
(a) An electron moves with a speed of 4.70 x 106 m/s. What is its de Broglie wavelength? (b) A proton moves with the same speed. Determine its de Broglie wavelength.
Answer:
λ = h/p = h/mv
For an electron, m = 9.11 x 10-31 kg. For a proton m = 1.67 x 10-27 kg.
(a) de Broglie wavelength is
λ = (6.63 x 10-34 J.s)/[(9.11 x 10-31 kg)(4.70 x 106 m/s)]
λ = 1.55 x 10-10 = 0.155 nm
(b) λ is proportional to 1/m,
so λp = λe(me/mp) = (1.55 x 10-10 m)(9.11 x 10-31 kg/1.67 x 10-27 kg) = 8.46 x 10-14 m.
Problem#2
For crystal diffraction experiments (discussed in Section 39.1), wavelengths on the order of 0.20 nm are often appropriate. Find the energy in electron volts for a particle with this wavelength if the particle is (a) a photon; (b) an electron; (c) an alpha particle (m = 6.64 x 10-27 kg).
Answer:
For a photon, E = hc/λ.
For an electron or proton, p = h/λ and E = p2/2m, so E = h2/2mλ2.
(a) E = hc/λ = (6.63 x 10-34 J.s)(3.00 x 108 m/s)/(0.20 x 10-9 m) = 6.2 keV
(b) E = h2/2mλ2 = (6.63 x 10-34 J.s)2/[2(9.11 x 10-31 kg)(0.20 x 10-9 m)] = 6.02 x 10-18 J = 38 eV
(c) Ep = Ee(me/mp) = (38 eV)(9.11 x 10-31 kg/1.67 x 10-27 kg) = 0.021 eV.
Problem#3
An electron has a de Broglie wavelength of 2.80 x 10-10 m. Determine (a) the magnitude of its momentum and (b) its kinetic energy (in joules and in electron volts).
Answer:
1 eV = 1.60 x 10-19 J
(a) λ = h/p → p = h/λ = (6.62 x 10-34 J.s)/(2.80 x 10-10 m) = 2.37 x 10-24 kg.m/s
(b) K = p2/2m = (2.37 x 10-24 kg.m/s)2/[2(9.11 x 10-31 kg)] = 3.08 x 10-18 J = 19.3 eV
Problem#4
Wavelength of an Alpha Particle. An alpha particle (m = 6.64 x 10-27 kg) emitted in the radioactive decay of uranium-238 has an energy of 4.20 MeV. What is its de Broglie wavelength?
Answer:
For a particle with mass, λ = h/p and E = p2/2m.
1 eV = 1.60 x 10-19 J
λ = h/p = h/√(2mE)
λ = (6.62 x 10-34 J.s)/√[2(6.64 x 10-27 kg)(4.20 x 106 eV)(1.60 x 10-19 J/eV)]
λ = 7.02 x 10-15 m
Problem#5
In the Bohr model of the hydrogen atom, what is the de Broglie wavelength for the electron when it is in (a) the n = 1 level and (b) the level? In each case, compare the de Broglie wavelength to the circumference 2πrn of the orbit.
Answer:
The de Bloglie wavelength is λ = h/p = h/mv. In the Bohr model
mvrn = n(h/2π), so
mv = nh/(2πrn)
Combine these two expressions and obtain an equation for λ in term of n. Then
λ = h[2πrn/nh] = 2πrn/n
(a) For n = 1, λ = 2πr1 with r1 = a0 = 0.529 x 10-10 m, so
λ = 2π(0.529 x 10-10 m) = 3.32 x 10-10 m
λ = 2πr1; the de Broglie wavelengt equals the circumference of the orbit.
(b) For n = 4, λ = 2πr4/4.
rn = n2a0, so r4 = 16a0
λ = 2π(16a0)/4 = 4 x (2πa0)
λ = 4 x 3.32 x 10-10 m = 1.33 x 10-9 m
λ = 2π/4; the de Broglie wavelength is 1/n = 1/4 times the circumference of the orbit.
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