Home / E.3 Radioactive decay SL Paper 2- IBDP Physics 2025- Exam Style Questions

E.3 Radioactive decay SL Paper 2- IBDP Physics 2025- Exam Style Questions

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]

E.3 Radioactive decay SL  Paper 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 ) \(\begin{aligned} & \lambda=\ll \frac{\ln 2}{12.3} \rightsquigarrow 0.056 \kappa y^{-1} » \text { OR } 0.5^{\frac{1}{m}} \text { OR } e^{-1 \times \frac{\ln 2}{12.3}} \\ & 0.945 \text { OR } 94.5 \%\end{aligned}\)

Question

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.

[2]
b.

Identify the missing entries to complete the nuclear reaction for the decay of I-131.

[1]
c.i.

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.

[2]
c.ii.

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.

[2]
 
Answer/Explanation

Markscheme

a.

same number of protons / atoms of the same element;

different number of neutrons;

b.

54 and antineutrino/\(\bar \nu \); (both needed)

c.i.

range is 14 to 26 or 14 to 27;

12 or 13 days;

Award [2] if marking points added to the graph.

c.ii.

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.

b.

Many seemed unaware of the antineutrino.

c.i.

Many candidates were able to determine the answer from the activity graph, but quite a few misunderstood what the question was asking for.

c.ii.

 

Question

Part 2 Radioactive decay

a.

Describe the phenomenon of natural radioactive decay.

[3]
b.

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).

[7]
 
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;

b.

(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.

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