IB Physics SL Theme E: Nuclear and quantum physics questions

Exam-style multiple choice and written questions with full markschemes. Check multiple choice answers straight away, or type, photograph or draw a written answer and get it marked.

Question 1

Multiple choiceEasy1 mark

A nucleus is represented by ZAX{}^{A}_{Z}X A second nucleus has the same proton number but one more neutron. What is its nuclear notation?

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Answer: A. AA is the nucleon number and ZZ is the proton number. One additional neutron increases the nucleon number by 1 while leaving the proton number, and therefore the element XX unchanged. The notation is ZA+1X{}^{A+1}_{Z}X

B: increases the proton number while keeping the nucleon number unchanged. C: decreases the nucleon number, corresponding to one fewer neutron at the same proton number. D: increases both the proton and nucleon numbers, adding a proton as well as a neutron.

Remember: adding a neutron changes AA not ZZ

Question 2

Paper 2Easy6 marks

Figure 1 shows nuclear notation and schematic emission and absorption spectra for an element.

2311Xemission spectrumabsorption spectrum400450500550wavelength / nm
(a)

In the nuclear notation shown in Figure 1, state what AA and ZZ represent, and calculate the number of neutrons in the nucleus.

[3]
(b)

Describe one feature of the line positions in the spectra shown in Figure 1.

[1]
(c)

Explain how the spectra provide evidence that atomic energy levels are discrete.

[2]

Question 3

Multiple choiceMedium1 mark

An atom emits a photon during the transition shown in the energy-level diagram. Take h=6.6×10−34 J sh=6.6\times10^{-34}\,\text{J s} and c=3.0×108 m s−1c=3.0\times10^8\,\text{m s}^{-1} What is the wavelength of the emitted photon?

Energy, E / J5.0×10−19 J4.0×10−19 J3.0×10−19 J2.0×10−19 J1.0×10−19 J0 J0 J1.1×10−19 J4.4×10−19 J
Show answer and explanation

Answer: C. The photon energy equals the difference between the atomic energy levels. Using E=hfE=hf and c=fλc=f\lambda λ=hc/ΔE\lambda=hc/\Delta E

  1. ΔE=(4.4−1.1)×10−19=3.3×10−19 J\Delta E=(4.4-1.1)\times10^{-19}=3.3\times10^{-19}\,\text{J}
  2. λ=(6.6×10−34)(3.0×108)3.3×10−19=6.0×10−7 m\lambda=\frac{(6.6\times10^{-34})(3.0\times10^8)}{3.3\times10^{-19}}=6.0\times10^{-7}\,\text{m}

A: adds the two level energies instead of finding their difference.
B: uses the upper level’s energy relative to zero rather than the energy gap shown.
D: uses the lower level’s energy as the photon energy rather than the difference between the levels.

Remember: a spectral line corresponds to a photon with energy equal to the difference between two discrete atomic energy levels.

Question 4

Multiple choiceEasy1 mark

For each nucleus in a sample of 500 radioactive nuclei, the probability of decay during a specified interval is 0.20. What is the expected number of nuclei that decay during the interval?

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Answer: C. The expected number of decays is the probability of decay per nucleus multiplied by the number of nuclei: 0.20×500=1000.20 \times 500 = 100 nuclei. Because decay is random, this is an expected number, not a guarantee of the exact number that will decay.

A: converts the probability 0.20 to 20% and reports 20 as the number of nuclei, without applying the probability to the sample of 500.
B: uses half the stated probability, 0.10, giving 50 nuclei.
D: calculates the expected number that do not decay, 0.80×500=4000.80 \times 500 = 400 nuclei.
Remember: probability predicts an expected fraction of a sample, not which individual nuclei decay.

Question 5

Paper 2Medium8 marks

Figure 1 shows four nuclear transformations. In the first sequence, the nucleus formed after alpha decay becomes the parent nucleus in the following beta-minus decay. The symbol * denotes an excited nucleus.

23892Uαnucleus formedin (a)β⁻23491Pa189Fβ⁺188O9943Tc*γ9943Tc
(a)

Write the complete nuclear equation for the alpha decay in the first sequence.

[3]
(b)

Using your answer to (a), write the complete nuclear equation for the beta-minus decay in the first sequence. Include all emitted particles.

[2]
(c)

Write the complete nuclear equation for the beta-plus decay shown in Figure 1. Include all emitted particles.

[2]
(d)

Write the complete nuclear equation for the gamma emission shown in Figure 1.

[1]

Question 6

Paper 2Easy9 marks

One possible neutron-induced fission reaction of uranium-235 is

²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + 3 ¹₀n

The following rest masses are available:

Particle or atomRest mass / u
²³⁵U235.043930
¹n1.008665
¹⁴¹Ba140.914411
⁹²Kr91.926156
(a)

Distinguish between spontaneous fission and neutron-induced fission.

[2]
(b)

For the reaction and data above, use 1 u c² = 931.5 MeV and 1 MeV = 1.60 × 10⁻¹³ J.

(i)

Determine the energy released in the reaction, in MeV.

[3]
(ii)

Using your answer to (b)(i), express the energy released in joules.

[2]
(c)

Explain how neutrons released in fission can sustain a chain reaction in uranium-235.

[2]

Question 7

Multiple choiceMedium1 mark

In a fission power plant, the fission rate has risen above its intended value. Fuel removed after use contains fission products. Which paired action correctly addresses the reactor rate and the removed fuel?

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Answer: D. Control rods absorb neutrons, so inserting them further reduces the number available to sustain fission and lowers the fission rate. Fission products in removed fuel are radioactive, so the fuel requires long-term storage.
A: confuses shielding, which limits radiation escaping the reactor, with control of the neutron population driving fission.
B: correctly controls the rate, but shielding does not remove the need for long-term storage of radioactive fuel.
C: withdrawing control rods absorbs fewer neutrons, tending to increase rather than decrease the fission rate.
Remember: control rods regulate fission; shielding limits radiation exposure, while radioactive waste needs long-term management.

Question 8

Multiple choiceMedium1 mark

A reactor operator is planning long-term storage for fission products, some of which have long half-lives. Which statement correctly links their nuclear behaviour to the need for this storage?

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Answer: B. A long half-life means the activity of a radioactive sample decreases slowly, so some fission products can remain radioactive over long periods. Secure long-term containment reduces exposure to radiation and prevents radioactive material from reaching the environment.

A: confuses shielding, which reduces radiation reaching people, with changing unstable nuclei into stable nuclei. C: assumes that stopping fission also stops radioactive decay; fission products continue to decay after removal. D: treats a half-life as a fixed expiry time for every nucleus; after one half-life, about half the radioactive nuclei are expected to have decayed.

Remember: shielding reduces exposure; it does not stop radioactive decay.

Question 9

Multiple choiceMedium1 mark

An HR diagram for stars P and Q is shown. Their unequal stellar parallaxes, measured in arc-seconds, are pPp_P and pQp_Q respectively. Which expression gives dred giant/dwhite dwarfd_{\text{red giant}}/d_{\text{white dwarf}}

252015105010-310-210-11101001000Surface temperature, T / 103 KLuminosity, L/LsunPQ
Show answer and explanation

Answer: C. On the HR diagram, P is cool and luminous, so it is a red giant; Q is hot and faint, so it is a white dwarf. Using d (pc)=1/p (arc-second)d\text{ (pc)}=1/p\text{ (arc-second)}

dred giantdwhite dwarf=1/pP1/pQ=pQpP.\frac{d_{\text{red giant}}}{d_{\text{white dwarf}}} =\frac{1/p_P}{1/p_Q} =\frac{p_Q}{p_P}.

A: Uses the parallax ratio directly, as though distance were proportional to parallax.
B: Reverses the required ratio and adds an unnecessary square.
D: Squares the correct inverse-parallax ratio.

Remember: distance and parallax are inversely proportional.

Question 10

Paper 2Medium7 marks

In the core of a young star, deuterium nuclei fuse with protons according to

12H+11H→23He+γ.{}^{2}_{1}\mathrm{H}+{}^{1}_{1}\mathrm{H}\rightarrow{}^{3}_{2}\mathrm{He}+\gamma .

Figure 1 shows the binding energy per nucleon for the nuclei relevant to this reaction. The plotted values are approximate.

01234012345678Nucleon number, ABinding energy per nucleon / MeV11H21H32He42He

Use 1 MeV=1.60×10−13 J1\ \mathrm{MeV}=1.60\times10^{-13}\ \mathrm{J}

(a)

Use Figure 1 to determine the energy released per reaction. Give your answer in MeV to two significant figures.

[3]
(b)

At a particular stage, this reaction occurs at a rate of 2.0×1037 s−12.0\times10^{37}\ \mathrm{s^{-1}} Calculate the energy released per second by these reactions. Give your answer in W to two significant figures.

[2]
(c)

The reaction rate then falls to half its value in (b), while the energy released per reaction remains unchanged. Determine the new energy released per second and explain how it compares with your answer to (b).

[2]

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