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Wave-particle duality questions
The photoelectric effect caught light behaving as particles. Electron diffraction catches matter behaving as waves. Neither label survives on its own, and one small equation, lambda equals h over mv, connects the two worlds.
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) = 5.52 × 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.
Explain what the diffraction of electrons by a thin crystal shows about the nature of electrons.
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).
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).
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).
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.
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).
Explain how and why the electron diffraction pattern changes when the electrons are accelerated to a higher speed.
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).
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).
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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