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Quanta and wave-particle duality questions
Two results broke the wave theory of light just after it had won. The ultraviolet catastrophe, in which classical physics predicted infinite energy from a hot body, and the photoelectric effect, in which bright red light ejects no electrons at all, both required light to be quantised. De Broglie then showed matter has a wavelength too.
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State what is meant by the ultraviolet catastrophe.
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Classical wave theory predicts that a black body's radiated intensity should rise without limit at short wavelengths, towards the ultraviolet (1), in complete contradiction of the measured curve, which peaks and falls (1).State Planck's interpretation of black-body radiation, and how it removes the catastrophe.
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Energy is emitted and absorbed in quanta of size E = hf (1). High-frequency quanta are expensive, so the short-wavelength modes are starved of energy and the curve turns over as observed (1).Give two observations about photoelectricity that classical wave theory cannot explain.
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Any two of the following, one mark each (2): the existence of a threshold frequency; emission is immediate, with no time to accumulate wave energy; increasing intensity raises the number of electrons but never their maximum kinetic energy.Write down the de Broglie relation between the momentum of a particle and its wavelength.
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p = h/λ (equivalently λ = h/p), where h is the Planck constant (1).Using the photon model, explain why increasing the intensity of the light falling on a photoemissive surface increases the number of electrons emitted each second but not their maximum kinetic energy.
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One photon interacts with one electron, giving it all of its energy hf (1). Greater intensity means more photons per second, so more electrons per second, but each photon still carries the same energy hf, so the maximum kinetic energy is unchanged (1).Calculate the de Broglie wavelength of electrons accelerated from rest through 150 V.
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λ = h/√(2meV) (1)
λ = 6.63 × 10−34/√(2 × 9.11 × 10−31 × 1.60 × 10−19 × 150) (1)
λ = 1.0 × 10−10 m (1)Calculate the de Broglie wavelength for an accelerating pd of 5.0 kV.
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λ = h/√(2meV) = 6.63 × 10−34/√(2 × 9.11 × 10−31 × 1.60 × 10−19 × 5000) (1)
λ = 1.7 × 10−11 m (1)In a low-energy electron diffraction experiment the accelerating pd is made nine times larger. State and explain the effect on the ring pattern.
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λ ∝ 1/√V, so the wavelength falls to a third of its value (1), and the diffraction rings shrink to a third of their radius (1): faster electrons, more momentum, shorter matter waves, tighter pattern (1).Explain the significance of Einstein's explanation of the photoelectric effect for the nature of electromagnetic radiation.
An ultraviolet lamp emits radiation of frequency 1.5 × 1015 Hz. Calculate the energy of one quantum of this radiation. h = 6.63 × 10−34 J s.
In an electron microscope column an electron travels at 4.0 × 106 m s−1. Calculate its de Broglie wavelength. h = 6.63 × 10−34 J s; me = 9.11 × 10−31 kg.
Electrons are accelerated from rest through 48 V towards a thin metal crystal in which the atomic planes are about 2 × 10−10 m apart. Show that the de Broglie wavelength of the electrons is about 1.8 × 10−10 m, and explain why a diffraction pattern is observed. h = 6.63 × 10−34 J s; me = 9.11 × 10−31 kg; e = 1.60 × 10−19 C.
Calculate the anode voltage that gives electrons a de Broglie wavelength of 5.0 × 10−11 m, half the size of a typical atom.
Describe the principle of operation of the transmission electron microscope.
Describe the principle of operation of the scanning tunnelling microscope.
A virus particle 90 nm across is to be imaged. The laboratory has an optical microscope using light of wavelength 500 nm and a simple electron microscope with an anode pd of 1.2 kV. A microscope cannot resolve detail much smaller than the wavelength it uses. Deduce which instrument can resolve the virus. h = 6.63 × 10−34 J s; me = 9.11 × 10−31 kg; e = 1.60 × 10−19 C.
Estimate the de Broglie wavelength of a 60 kg sprinter running at 10 m s−1, and hence explain why the wave nature of matter is never noticed for everyday objects. h = 6.63 × 10−34 J s.
The photoelectric effect could not be explained by the classical wave theory of light. Explain fully how the observations contradict the wave theory, and how Einstein's photon model accounts for them.
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