Questions on quanta and the photoelectric effect

Question 1

State what is meant by the ultraviolet catastrophe.

Accepted Answer

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).

Question 2

State Planck's interpretation of black-body radiation, and how it removes the catastrophe.

Accepted Answer

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).

Question 3

Give two observations about photoelectricity that classical wave theory cannot explain.

Accepted Answer

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.

Question 4

Write down the de Broglie relation between the momentum of a particle and its wavelength.

Accepted Answer

p = h/λ (equivalently λ = h/p), where h is the Planck constant (1).

Question 5

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.

Accepted Answer

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).

Question 6

Calculate the de Broglie wavelength of electrons accelerated from rest through 150 V.

Accepted Answer

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)

Question 7

Calculate the de Broglie wavelength for an accelerating pd of 5.0 kV.

Accepted Answer

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)

Question 8

In a low-energy electron diffraction experiment the accelerating pd is made nine times larger. State and explain the effect on the ring pattern.

Accepted Answer

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).

Question 9

Explain the significance of Einstein's explanation of the photoelectric effect for the nature of electromagnetic radiation.

Accepted Answer

Einstein took light itself to travel as quanta, photons of energy hf (1), one photon releasing one electron (1). Its success showed that electromagnetic radiation, so recently established as a wave, also behaves as particles: wave-particle duality for light (1).

Question 10

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.

Accepted Answer

E = hf (1) = 6.63 × 10⁻³⁴ × 1.5 × 10¹⁵ = 9.9 × 10⁻¹⁹ J (1).

Question 11

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; m_e = 9.11 × 10⁻³¹ kg.

Accepted Answer

p = mv = 9.11 × 10⁻³¹ × 4.0 × 10⁶ = 3.6 × 10⁻²⁴ kg m s⁻¹ (1). λ = h/p (1) = 6.63 × 10⁻³⁴/(3.6 × 10⁻²⁴) = 1.8 × 10⁻¹⁰ m (1).

Question 12

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; m_e = 9.11 × 10⁻³¹ kg; e = 1.60 × 10⁻¹⁹ C.

Accepted Answer

λ = h/√(2meV) (1) = 6.63 × 10⁻³⁴/√(2 × 9.11 × 10⁻³¹ × 1.60 × 10⁻¹⁹ × 48) = 1.77 × 10⁻¹⁰ m, about 1.8 × 10⁻¹⁰ m (1). This wavelength is comparable to the spacing of the atomic planes (1), so the regular planes act as a diffraction grating for the electron waves and a diffraction pattern appears (1).

Question 13

Calculate the anode voltage that gives electrons a de Broglie wavelength of 5.0 × 10⁻¹¹ m, half the size of a typical atom.

Accepted Answer

λ = h/√(2meV) (1) rearranges to V = h²/(2meλ²) (1) V = (6.63 × 10⁻³⁴)²/(2 × 9.11 × 10⁻³¹ × 1.60 × 10⁻¹⁹ × (5.0 × 10⁻¹¹)²) (1) V = 603 V (1)

Question 14

Describe the principle of operation of the transmission electron microscope.

Accepted Answer

An electron gun fires a beam down an evacuated column (1); magnetic coils act as lenses, condensing the beam through a very thin specimen and projecting the transmitted electrons onto a fluorescent screen (1). Denser regions scatter more electrons and appear darker, and the short de Broglie wavelength allows detail far beyond any light microscope (1).

Question 15

Describe the principle of operation of the scanning tunnelling microscope.

Accepted Answer

A tip sharpened almost to a single atom is held about a nanometre above a conducting surface (1), close enough for electrons to tunnel across the gap (1). The tunnelling current falls extremely steeply with distance, so scanning the tip across the surface and monitoring the current maps its atoms (1).

Question 16

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; m_e = 9.11 × 10⁻³¹ kg; e = 1.60 × 10⁻¹⁹ C.

Accepted Answer

The light's 500 nm wavelength is larger than the 90 nm virus, so the optical microscope cannot resolve it (1). Electrons: λ = h/√(2meV) = 6.63 × 10⁻³⁴/√(2 × 9.11 × 10⁻³¹ × 1.60 × 10⁻¹⁹ × 1200) (1) = 3.5 × 10⁻¹¹ m, about 0.035 nm (1). This is far smaller than 90 nm, so the electron microscope resolves the virus easily (1).

Question 17

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.

Accepted Answer

p = mv = 60 × 10 = 600 kg m s⁻¹ (1). λ = h/p = 6.63 × 10⁻³⁴/600 ≈ 1.1 × 10⁻³⁶ m (1). No gap or obstacle is remotely as small as 10⁻³⁶ m, so everyday objects can never show measurable diffraction (1).

Question 18

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.

Accepted Answer

Wave theory lets energy arrive continuously and accumulate, so light of any frequency should eventually eject electrons; in fact no electrons are emitted below a threshold frequency (1). Wave theory predicts a delay while a dim source's energy builds up; in fact emission is immediate at any intensity (1). Wave theory predicts brighter light should give each electron more energy; in fact intensity changes the number emitted, never the maximum kinetic energy (1). Photon model: light travels as photons of energy hf, and one photon gives all its energy to one electron (1). Emission happens only if hf exceeds the energy needed to escape the surface, giving the threshold frequency and instant emission (1). Greater intensity means more photons and so more electrons, while the maximum kinetic energy depends only on hf (1).