Theme E · Nuclear and quantum physics

An electron in a hydrogen atom moves from an energy level of −1.51 eV to a level of −3.40 eV. What is the wavelength of the emitted photon? (hc = 1240 eV nm)

1. A254 nm
2. B365 nm
3. C656 nm
4. D821 nm

A nuclide undergoes beta-minus decay followed by alpha decay. Compared with the original nuclide, the final nuclide has:

1. A4 fewer nucleons and 1 fewer proton
2. B4 fewer nucleons and 2 fewer protons
3. C4 fewer nucleons and 3 fewer protons
4. D2 fewer nucleons and 1 fewer proton

A sample contains 8.0 × 10²⁰ nuclei of a radioactive isotope with a half-life of 12 days. How many nuclei remain undecayed after 36 days?

1. A1.0 × 10²⁰
2. B2.0 × 10²⁰
3. C2.7 × 10²⁰
4. D4.0 × 10²⁰

Which statement about the binding energy per nucleon curve is correct?

1. AIt rises steadily with nucleon number across the whole range.
2. BIt peaks near iron, so both fusing lighter nuclei and splitting heavier nuclei release energy.
3. CIt peaks near uranium, which is why uranium is used as a nuclear fuel.
4. DIt is the energy released when a nucleus is formed from its nucleons, divided by the nucleon number, and is greatest for hydrogen.

Two stars have the same surface temperature, but star P has four times the radius of star Q. What is the ratio of their luminosities, L_P : L_Q?

1. A4 : 1
2. B8 : 1
3. C16 : 1
4. D256 : 1

In a photoelectric experiment, the intensity of the incident light is increased while its frequency is held constant. What happens to the maximum kinetic energy of the emitted photoelectrons and to the photocurrent?

1. Aboth increase
2. Bmaximum kinetic energy unchanged, photocurrent increases
3. Cmaximum kinetic energy increases, photocurrent unchanged
4. Dboth unchanged

A student measures the activity A of a radioactive source at intervals, correcting each reading for background. She expects A = A₀e^(−λt).

1. (a)

   Explain Explain why a graph of ln A against t is plotted, and state what its gradient represents.

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   1. 1

      taking natural logarithms of A = A₀e^(−λt) gives ln A = ln A₀ − λt

   2. 2

      this is linear in t, so the points lie on a straight line if the decay is exponential — which the plot therefore also tests

   3. 3

      the gradient is −λ, the negative of the decay constant

2. (b)

   Determine Determine the decay constant and hence the half-life of the source.

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   1. 1

      gradient = (5.288 − 6.215)/(40 − 0) = −0.0232 min⁻¹

   2. 2

      λ = 0.0232 min⁻¹

   3. 3

      t½ = ln 2 / λ = 0.693/0.0232

   4. 4

      t½ = 30 minutes

3. (c)

   Outline Outline why the readings had to be corrected for background before being processed.

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   1. 1

      the detector registers radiation from cosmic rays, rocks and other natural sources as well as from the sample

   2. 2

      background adds a constant to every reading, which is a systematic error — it does not decay away, so an uncorrected plot of ln A against t would curve rather than being straight

4. (d)

   Explain The student's individual readings scatter about the line even after correction. Explain why this scatter cannot be removed by taking more care with the apparatus.

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   1. 1

      radioactive decay is a random process — which nucleus decays, and when, cannot be predicted

   2. 2

      so the number of decays counted in any fixed interval fluctuates about a mean value, and this fluctuation is a property of the process, not of the instrument

   3. 3

      it can only be reduced by counting for longer or over more nuclei, since the fractional fluctuation falls as the total count rises — not by improving the apparatus

The nuclide ²²⁶₈₈Ra decays by alpha emission with a half-life of 1600 years. A sealed sample initially contains 3.0 × 10²¹ radium-226 nuclei.

1. (a)

   State State the nucleon number and the proton number of the nuclide produced by this decay.

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   1. 1

      Alpha emission reduces the nucleon number by four, giving 222.

   2. 2

      Alpha emission reduces the proton number by two, giving 86.

2. (b)

   Determine Determine the initial activity of the sample, in becquerel.

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   1. 1

      t½ = 1600 years = 1600 × 3.15 × 10⁷ = 5.04 × 10¹⁰ s

   2. 2

      λ = ln 2 / t½ = 0.693 / 5.04 × 10¹⁰ = 1.37 × 10⁻¹¹ s⁻¹

   3. 3

      A = λN = 1.37 × 10⁻¹¹ × 3.0 × 10²¹

   4. 4

      A = 4.1 × 10¹⁰ Bq

3. (c)

   Explain Explain what is meant by saying that radioactive decay is both random and spontaneous.

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   1. 1

      random: it cannot be predicted which nucleus will decay next, or when a particular nucleus will decay — only the probability per unit time is known

   2. 2

      spontaneous: the decay is unaffected by external conditions such as temperature, pressure or chemical state

4. (d)

   Explain Explain why the emission spectrum of a gas consists of discrete lines rather than a continuous range of wavelengths, and outline what this shows about the atom.

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   1. 1

      an atom emits a photon when an electron moves from a higher to a lower energy level

   2. 2

      the photon energy equals the difference between the two levels, and its wavelength follows from E = hc/λ

   3. 3

      only certain transitions are possible, so only certain photon energies — and hence only certain wavelengths — are emitted

   4. 4

      this shows that the energy of an electron in an atom is quantised, taking only discrete values rather than any value in a range

In one fission event, a nucleus of ²³⁵U absorbs a neutron and splits, releasing about 200 MeV of energy. In the core of the Sun, four hydrogen nuclei are effectively converted into one helium-4 nucleus, releasing about 26 MeV.

1. (a)

   Determine Determine the energy released, in joules, per kilogram of uranium-235 undergoing fission. The mass of a ²³⁵U atom may be taken as 3.9 × 10⁻²⁵ kg.

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   1. 1

      number of nuclei per kg = 1 / 3.9 × 10⁻²⁵ = 2.56 × 10²⁴

   2. 2

      energy per fission = 200 × 10⁶ × 1.60 × 10⁻¹⁹ = 3.2 × 10⁻¹¹ J

   3. 3

      energy per kg = 2.56 × 10²⁴ × 3.2 × 10⁻¹¹

   4. 4

      = 8.2 × 10¹³ J kg⁻¹

2. (b)

   Explain Explain, in terms of binding energy per nucleon, why both fission of uranium and fusion of hydrogen release energy.

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   1. 1

      energy is released whenever the products are more tightly bound — have a greater binding energy per nucleon — than the reactants

   2. 2

      uranium lies to the right of the peak of the curve, so its fragments are closer to iron and more tightly bound

   3. 3

      hydrogen lies far to the left, so helium formed from it is much more tightly bound

3. (c)

   Explain Explain why extremely high temperatures are required for fusion but not for fission.

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   1. 1

      fusing nuclei are both positively charged and must be brought within range of the strong nuclear force

   2. 2

      this requires them to overcome a large electrostatic repulsion, which needs very high kinetic energies and hence very high temperatures

   3. 3

      a neutron has no charge, so it experiences no repulsion and can enter a uranium nucleus at ordinary — even low — energies

4. (d)

   Discuss The Sun's luminosity is 3.8 × 10²⁶ W. Discuss whether the fusion of hydrogen can account for the Sun's output over its estimated remaining lifetime of about 5 × 10⁹ years.

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   1. 1

      total energy required = 3.8 × 10²⁶ × 5 × 10⁹ × 3.15 × 10⁷ = 6.0 × 10⁴³ J

   2. 2

      each helium nucleus formed releases 26 MeV = 4.2 × 10⁻¹² J, using about 6.7 × 10⁻²⁷ kg of hydrogen

   3. 3

      so the mass of hydrogen required is about 6.0 × 10⁴³ × 6.7 × 10⁻²⁷ / 4.2 × 10⁻¹² ≈ 1 × 10²⁹ kg

   4. 4

      this is around 5% of the Sun's mass of 2 × 10³⁰ kg, and since fusion occurs only in the hot dense core rather than throughout the Sun, this is consistent with the estimated lifetime

In a photoelectric experiment, light of wavelength 420 nm is incident on a caesium surface of work function 2.1 eV. Planck's constant is 6.63 × 10⁻³⁴ J s and hc may be taken as 1240 eV nm.

1. (a)

   Determine Determine the maximum kinetic energy of the emitted photoelectrons, in electronvolts.

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   1. 1

      photon energy = 1240/420 = 2.95 eV

   2. 2

      E_k(max) = hf − φ = 2.95 − 2.1

   3. 3

      E_k(max) = 0.85 eV

2. (b)

   Determine Determine the threshold wavelength for caesium.

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   1. 1

      at threshold hf = φ, so λ_max = 1240/2.1

   2. 2

      λ_max = 590 nm

3. (c)

   Determine Determine the de Broglie wavelength of an electron emitted with the maximum kinetic energy found in (a).

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   1. 1

      E_k = 0.85 × 1.60 × 10⁻¹⁹ = 1.36 × 10⁻¹⁹ J

   2. 2

      p = √(2mE_k) = √(2 × 9.11 × 10⁻³¹ × 1.36 × 10⁻¹⁹)

   3. 3

      p = 4.98 × 10⁻²⁵ kg m s⁻¹

   4. 4

      λ = h/p = 6.63 × 10⁻³⁴ / 4.98 × 10⁻²⁵ = 1.3 × 10⁻⁹ m

4. (d)

   Explain Explain how the existence of a threshold frequency, and the absence of any measurable time delay before emission begins even in very dim light, together provide evidence against a purely wave model of light.

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   1. 1

      a wave model predicts that energy arrives continuously and spreads over the whole surface, so any frequency should eventually free an electron if the light shines long enough

   2. 2

      instead, no electrons are emitted below the threshold frequency however intense or prolonged the illumination — so the energy must arrive in single indivisible amounts of size hf

   3. 3

      a wave model also predicts a measurable delay in very dim light while an electron accumulates enough energy

   4. 4

      no such delay is observed, because a single photon delivers all of its energy to a single electron in one interaction