Problem#1
(a) A nonrelativistic free particle with mass m has kinetic energy K. Derive an expression for the de Broglie wavelength of the particle in terms of m and K. (b) What is the de Broglie wavelength of an 800-eV electron?
Answer:
λ = h/p
1 ev = 1.60 x 10-19 J. An electron has mass 9.11 x 10-31 kg.
(a) For a nonrelativiastic particle, K = p2/2m, so λ = h/p = h/√(2mK)
(b) λ = (6.62 x 10-34 J.s)/√[2(800eV x 1.60 x 10-10 J/eV)(9.11 x 10-31 kg)]
λ = 4.34 x 10-11 m
Problem#2
Why Don’t We Diffract? (a) Calculate the de Broglie wavelength of a typical person walking through a doorway. Make reasonable approximations for the necessary quantities. (b) Will the person in part (a) exhibit wavelike behavior when walking through the “single slit” of a doorway? Why?
Answer:
A person walking through a door is like a particle going through a slit and hence should exhibit wave properties.
The de Broglie wavelength of the person is λ = h/mv
(a) Assume m = 75 kg and v = 1.0 m/s.
λ = h/mv = (6.62 x 10-34 J.s)/[75 kg x 1.0 m/s] = 8.8 x 10-36 m
(b) A typical doorway is about 1 m wide, so the person’s de Broglie wavelength is much too
small to show wave behavior through a “slit” that is about 1035 times as wide as the wavelength. Hence ordinary objects do not show wave behavior in everyday life.
Problem#3
What is the de Broglie wavelength for an electron with speed (a) v = 0.480c and (b) v = 0.960c (Hint: Use the correct relativistic expression for linear momentum if necessary.)
Answer:
Gives p = mc√[γ2 – 1]
(a) for v = 0.480c, then γ = [1 – v2/c2]-1 = 1.1399
λ = h/p = h/mc√[γ2 – 1]
λ = (6.63 x 10-34 J.s)/{(9.11 x 10-31 kg)(3.00 x 108 m/s)√[(1.1399)2 – 1]
λ = 4.43 x 10-12 m (The incorrect nonrelativistic calculation gives 5.05 x 10-12 m.
(b) for v = 0.960c, then γ = [1 – v2/c2]-1 = 3.571
λ = h/p = h/mc√[γ2 – 1]
λ = (6.63 x 10-34 J.s)/{(9.11 x 10-31 kg)(3.00 x 108 m/s)√[(3.571)2 – 1]
λ = 7.07 x 10-13 m
Problem#4
(a) If a photon and an electron each have the same energy of 20.0 eV, find the wavelength of each. (b) If a photon and an electron each have the same wavelength of 250 nm, find the energy
of each. (c) You want to study an organic molecule that is about 250 nm long using either a photon or an electron microscope. Approximately what wavelength should you use, and which probe, the electron or the photon, is likely to damage the molecule the least?
Answer:
A photon has zero mass and its energy and wavelength are related by Eq. (38.2). An electron has mass. Its energy is related to its momentum by E = p2/2m and its wavelength is related to its momentum by Eq. (39.1)
(a) Photon: E = hc/λ so
λ = hc/E = (6.626 x 10-34 J.s)(2.998 x 108 m/s)/[20.0 eV x 1.602 x 10-19 J/eV] = 62.0 nm
(b) Photon: E = hc/R = 7.946 x 10-19 J = 4.96 eV
Electron: λ = h/p so
p = h/λ = (6.626 x 10-34 J.s)/250 x 10-9 m = 2.650 x 10-27 kg.m/s
E = p2/2m = (2.650 x 10-27 kg.m/s)2/(2 x 9.11 x 10-31 kg) = 2.41 x 10-5 eV
(c) You should use a probe of wavelength approximately 250 nm. An electron with
λ = 250 nm has much less energy than a photon with 250 nm, λ = so is less likely to damage the
molecule. Note that λ = h/p applies to all particles, those with mass and those with zero mass.
E = hf = hc/λ applies only to photons and E = p2/2m applies only to particles with mass.
Problem#5
How fast would an electron have to move so that its de Broglie wavelength is 1.00 mm?
Answer:
Knowing the de Broglie wavelength for an electron, we want to find its speed
λ = h/p = h/mv
v = h/mλ = (6.63 x 10-34 J.s)/(9.11 x 10-31 kg x 1.00 x 10-3 m) = 0.728 m/s
Electrons normally move much faster than this, so their de Broglie wavelengths are much much smaller than a millimeter.
Problem#6
Wavelength of a Bullet. Calculate the de Broglie wavelength of a 5.00-g bullet that is moving at Will the bullet exhibit wavelike properties?
Answer:
λ = h/p = h/mv = (6.63 x 10-34 J.s)/(9.11 x 10-31 kg x 340 m/s) = 3.90 x 10-34 m
This wavelength is extremely short; the bullet will not exhibit wavelike properties.
Post a Comment for "Electron Waves Problems and Solutions 1"