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Questions on wave-particle duality
Electron diffraction through graphite shows particles behaving as waves, while the photoelectric effect shows light behaving as particles. The de Broglie wavelength λ = h/mv links the two, so raising a particle's momentum shortens its wavelength and tightens the diffraction rings. Everyday objects carry far too much momentum for their wave behaviour to be measurable.
19 original questions · 55 marks · the wave-particle duality notes · Quantum phenomena
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State the two pieces of evidence showing that light and electrons each have both wave and particle properties.
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Electron diffraction shows that electrons (particles) behave as waves (1); the photoelectric effect shows that light (a wave) behaves as particles (photons) (1).A particle has a momentum of 2.0 × 10−24 kg m s−1. Calculate its de Broglie wavelength (h = 6.63 × 10−34 J s).
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λ = h/p = (6.63 × 10−34)/(2.0 × 10−24) (1)
λ = 3.32 × 10−10 m (1)State the de Broglie equation and define each term.
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λ = h/mv (= h/p) (1), where λ is the de Broglie wavelength, h the Planck constant, m the mass, v the speed and p the momentum (1).The electrons in an electron microscope have a de Broglie wavelength of 1.2 × 10−10 m. Calculate their momentum (h = 6.63 × 10−34 J s).
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p = h/λ = (6.63 × 10−34)/(1.2 × 10−10) (1) = 5.53 × 10−24 kg m s−1 (1).State the condition needed for a beam of particles to be diffracted appreciably by the atoms of a crystal.
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The de Broglie wavelength must be comparable to the atomic spacing (1).The momentum of a particle is doubled. State what happens to its de Broglie wavelength.
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It halves, since λ = h/p (1).An electron moves at 1.0 × 106 m s−1. Calculate its de Broglie wavelength (me = 9.11 × 10−31 kg).
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λ = h/mv (1)
= (6.63 × 10−34)/(9.11 × 10−31 × 1.0 × 106) (1)
λ = 7.28 × 10−10 m (1)An electron is accelerated from rest through a potential difference of 100 V. Calculate (a) its speed and (b) its de Broglie wavelength (e = 1.60 × 10−19 C, me = 9.11 × 10−31 kg).
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Ek = eV = 1.60 × 10−17 J (1)
(a) v = √(2Ek/m) (1)
v = 5.93 × 106 m s−1 (1)
(b) λ = h/mv = 1.23 × 10−10 m (1)An electron and a proton travel at the same speed of 1.0 × 105 m s−1. Calculate the de Broglie wavelength of each (mp = 1.67 × 10−27 kg) and state which is larger.
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Electron: λ = h/mv = 7.28 × 10−9 m (1)
Proton: λ = 3.97 × 10−12 m (1)
The electron's wavelength is larger, because it has the smaller mass (1)Explain what the diffraction of electrons by a thin crystal shows about the nature of electrons.
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Diffraction is a wave property, produced when a wave passes through gaps comparable to its wavelength (1) (here the atomic spacing). Electrons producing a diffraction pattern therefore behave as waves (1), evidence for wave–particle duality (1).In an experiment to show that whole atoms have wave properties, a beam of helium atoms is directed at a crystal whose atomic spacing is 1.0 × 10−10 m. Calculate the speed at which a helium atom, of mass 6.6 × 10−27 kg, has a de Broglie wavelength equal to this spacing (h = 6.63 × 10−34 J s).
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λ = h/mv, so v = h/mλ (1) = (6.63 × 10−34)/(6.6 × 10−27 × 1.0 × 10−10) (1) = 1.00 × 103 m s−1, about 1.0 km s−1 (1).An electron travels at 3.3 × 106 m s−1. Determine the speed at which a proton would have the same de Broglie wavelength as this electron (me = 9.11 × 10−31 kg, mp = 1.67 × 10−27 kg).
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Equal wavelengths mean equal momenta: meve = mpvp (1). vp = (9.11 × 10−31 × 3.3 × 106)/(1.67 × 10−27) (1) = 1.80 × 103 m s−1 (1).In a molecular-beam experiment, molecules travelling at 500 m s−1 produce a diffraction pattern corresponding to a de Broglie wavelength of 2.8 × 10−11 m. The beam is one of: helium (particle mass 6.6 × 10−27 kg), nitrogen (4.7 × 10−26 kg) or oxygen (5.3 × 10−26 kg). Determine the mass of one molecule and deduce which gas the beam contains (h = 6.63 × 10−34 J s).
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λ = h/mv, so m = h/(λv) (1) = (6.63 × 10−34)/(2.8 × 10−11 × 500) (1) = 4.7 × 10−26 kg (1). This matches nitrogen, so the beam is nitrogen (1).A cricket ball of mass 0.16 kg is bowled at 30 m s−1. Calculate its de Broglie wavelength and explain why no diffraction of the ball is ever observed.
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λ = h/mv = (6.63 × 10−34)/(0.16 × 30) (1)
λ = 1.38 × 10−34 m (1)
This is many orders of magnitude smaller than any gap the ball could pass through (1)
so any diffraction is far too small to detect (1)Calculate the accelerating potential difference needed to give an electron a de Broglie wavelength of 1.0 × 10−10 m (h = 6.63 × 10−34 J s, me = 9.11 × 10−31 kg, e = 1.60 × 10−19 C).
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p = h/λ = (6.63 × 10−34)/(1.0 × 10−10) (1)
p = 6.63 × 10−24 kg m s−1 (1)
Ek = p2/2m = 2.41 × 10−17 J (1)
V = Ek/e = 151 V (1)Explain how and why the electron diffraction pattern changes when the electrons are accelerated to a higher speed.
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A higher speed means greater momentum, so from λ = h/mv the de Broglie wavelength is shorter (1). With a shorter wavelength the electrons diffract less (1), so the rings of the pattern move closer together (the pattern shrinks) (1).A researcher wants to study a crystal whose atomic plane spacing is 2.0 × 10−10 m by diffraction. Two beams are available: electrons moving at 1.0 × 107 m s−1, or neutrons moving at 2.0 × 103 m s−1. Deduce which beam is better matched to the crystal (h = 6.63 × 10−34 J s, me = 9.11 × 10−31 kg, neutron mass = 1.67 × 10−27 kg).
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Electrons: λ = h/mv = 7.28 × 10−11 m (1). Neutrons: λ = 1.99 × 10−10 m (1). Diffraction is strongest when λ is comparable to the spacing (1). The neutron wavelength is almost exactly 2.0 × 10−10 m, while the electron wavelength is nearly three times smaller, so the neutron beam is the better match (1).Show that the kinetic energy of a particle of mass m with de Broglie wavelength λ is given by Ek = h2/(2mλ2). Go on to calculate the kinetic energy of an electron whose de Broglie wavelength is 5.0 × 10−10 m (h = 6.63 × 10−34 J s, me = 9.11 × 10−31 kg).
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p = h/λ (1). Ek = p2/2m, so Ek = h2/(2mλ2) (1). Ek = (6.63 × 10−34)2/(2 × 9.11 × 10−31 × (5.0 × 10−10)2) (1) = 9.65 × 10−19 J, about 6.0 eV (1).The smallest detail a microscope can resolve is roughly the size of the wavelength it uses. Visible light has wavelengths around 5 × 10−7 m; the electrons in an electron microscope have a de Broglie wavelength of about 1 × 10−11 m. Explain why an electron microscope can resolve far finer detail than an optical microscope.
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Resolution is limited by diffraction, which blurs any detail smaller than about one wavelength (1). The electron de Broglie wavelength, about 10−11 m, is tens of thousands of times shorter than that of visible light (1), so diffraction only limits the electron microscope at a far smaller scale and it can resolve much finer detail (1).
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