IBDP Physics SL 2025 -E.3 Radioactive decay SL Paper 2 Exam Style Questions
IBDP Physics 2025 SL Paper 2 – All Chapters
Topic: E.3 Radioactive decay SL Paper 2
Nuclear Biding Energy, Mass-Energy Equivalence, Strong Nuclear Force, Alpha/Beta/Gamma Decay, Half-Life
Question-E.3 Radioactive decay SL Paper 2
(a) Identify with ticks [ \(\checkmark\) ] in the table, the forces that can act on electrons and the forces that can act on quarks. [2]
(b) The following data is available for atomic masses for the fusion reaction
$
{ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+{ }_0^1 \mathrm{n}:
$
(i) Show that the energy released is about 18MeV. [2]
(ii) Estimate the specific energy of hydrogen by finding the energy produced when \(0.4 \mathrm{~kg}\) of \({ }_1^2 \mathrm{H}\) and \(0.6 \mathrm{~kg}\) of \({ }_1^3 \mathrm{H}\) undergo fusion. [2]
(c) It is hoped that nuclear fusion can be used for commercial production of energy.
Outline
(i) two difficulties of energy production by nuclear fusion.[2]
(ii) one advantage of energy production by nuclear fusion compared to nuclear fission. [1]
(d) Tritium \(\left({ }_1^3 \mathrm{H}\right)\) is unstable and decays into an isotope of helium \((\mathrm{He})\) by beta minus decay with a half-life of 12.3 years.
(i) State the nucleon number of the \(\mathrm{He}\) isotope that \({ }_1^3 \mathrm{H}\) decays into.[1]
(ii) The following diagram is an incomplete Feynman diagram describing the beta minus decay of \({ }_1^3 \mathrm{H}\) into \(\mathrm{He}\). Complete the diagram and label all the missing particles.[3]
(iii) Estimate the fraction of tritium remaining after one year. [2]
▶️Answer/Explanation
Ans:
a ) Weak nuclear: 2 ticks \(\checkmark\)
Strong nuclear: quarks only \(\checkmark\)
b( i)
$
\varangle \mu »=2.0141+3.0160-(4.0026+1.008665) \ll=0.0188 u »
$
OR
In MeV: \(1876.13415+2809.404-(3728.4219+939.5714475)\)
$
=0.0188 \times 931.5 \text { OR }=17.512 \alpha \mathrm{MeV} » \checkmark
$
ALTERNATIVE 1
\(0.40 \mathrm{~kg}\) of deuterium is \(\ll \frac{400}{2} \times 6.02 \times 10^{23} \mathrm{w}=1.2 \times 10^{26}\) nuclei
\({ }_\alpha 0.60 \mathrm{~kg}\) of tritium is the same number \({ }_* \checkmark\)
So specific energy \(\alpha \frac{1.2 \times 10^{26} \times 17.51 \times 10^6 \times 1.6 \times 10^{-19}}{0.4+0.6} \rrbracket=3.4 \times 10^{14} \ll \mathrm{J} \mathrm{kg}^{-1} » \checkmark\)
ALTERNATIVE 2
\(\alpha 17.51 \times 10^6 \times 1.6 \times 10^{-19}=» 2.8 \times 10^{-12} \ll \mathrm{J} »\)
AND
$
\begin{aligned}
& \ll(2.0141+3.0160) \times 1.66 \times 10^{-27}=» 8.35 \times 10^{-27} \\
& \ll \frac{2.8 \times 10^{-12}}{8.35 \times 10^{-37}},=3.4 \times 10^{14} \ll \mathrm{Jkg}^{-1}
\end{aligned}
$
c i Requires high temp/pressure
Must overcome Coulomb/intermolecular repulsion
Difficult to contain / control «at high temp/pressurew
Difficult to produce excess energy/often energy input greater than output / OWTTE \(\checkmark\)
Difficult to capture energy from fusion reactions
Difficult to maintain/sustain a constant reaction rate
ii ) Plentiful fuel supplies \(O R\) larger specific energy \(O R\) larger energy density \(O R\) little or no «major radioactive» waste products
d i ). 3
ii ) Proton shown
W- shown
Produces electron \(/ \mathrm{e}^{-} / \beta^{-}\)and antineutrino \(/ \bar{v}\) with proper arrow directions.
iii ) \(\begin{aligned} & \lambda=\ll \frac{\ln 2}{12.3} \rightsquigarrow 0.056 \kappa y^{-1} » \text { OR } 0.5^{\frac{1}{m}} \text { OR } e^{-1 \times \frac{\ln 2}{12.3}} \\ & 0.945 \text { OR } 94.5 \%\end{aligned}\)
This question is about radioactive decay.
A nucleus of an iodine isotope, I-131, undergoes radioactive decay to form a nucleus of the nuclide xenon-131. Xe-131 is stable.
The initial activity of a sample of I-131 is 100 kBq. The subsequent variation of the activity of the sample with time is shown in the graph.
a.
Explain what is meant by an isotope.
Identify the missing entries to complete the nuclear reaction for the decay of I-131.
The I-131 can be used for a medical application but only when the activity lies within the range of \((20 \pm 10){\text{ kBq}}\). Determine an estimate for the time during which the iodine can be used.
A different isotope has half the initial activity and double the half-life of I-131. On the graph in (c), sketch the variation of activity with time for this isotope.
Answer/Explanation
Markscheme
a.
same number of protons / atoms of the same element;
different number of neutrons;
54 and antineutrino/\(\bar \nu \); (both needed)
range is 14 to 26 or 14 to 27;
12 or 13 days;
Award [2] if marking points added to the graph.
starts at 50 kBq and approximately exponential decay curve;
half-life is \( \sim {\text{16 days}}\) / line passes through \([16,{\text{ }}25]\) to within a small square;
Examiners report
a.
A variety of good answers.
Many seemed unaware of the antineutrino.
Many candidates were able to determine the answer from the activity graph, but quite a few misunderstood what the question was asking for.
Part 2 Radioactive decay
a.
Describe the phenomenon of natural radioactive decay.
A nucleus of americium-241 (Am-241) decays into a nucleus of neptunium-237 (Np-237) in the following reaction.
\[{}_{95}^{241}{\rm{Am}} \to {}_X^{237}{\rm{Np}} + {}_2^4\alpha \]
(i) State the value of X.
(ii) Explain in terms of mass why energy is released in the reaction in (b).
(iii) Define binding energy of a nucleus.
(iv) The following data are available.
Determine the energy released in the reaction in (b).
Answer/Explanation
Markscheme
a.
emission of (alpha/beta/gamma) particles/photons/electromagnetic radiation;
nucleus becomes more (energetically) stable;
constant probability of decay (per unit time);
is random process;
activity/number of unstable nuclei in sample reduces by half over constant time intervals/exponentially;
not affected by temperature/environment / is spontaneous process;
(i) 93;
(ii) mass of products is less than mass of reactants / there is a mass defect;
mass is converted into energy (according to equation E=mc2);
(iii) the (minimum) energy required to (completely) separate the nucleons in a nucleus / the energy released when a nucleus is assembled from its constituent nucleons;
(iv) calculation of binding energies as shown below;
americium-241 = 241×7.54=1817.14 MeV
neptunium-237 = 237×7.58 = 1796.46 MeV
helium-4 = 4×7.07 = 28.28 MeV
energy released is the difference of binding energies;
and so equals 7.60 MeV;
Award [2 max] for an answer that multiplies by the number of neutrons or number of protons.