# E.5 Fusion SL Paper 2- IBDP Physics 2025- Exam Style Questions

IBDP Physics SL 2025 -E.5 Fusion SL Paper 2 Exam Style Questions

## Topic:E.5 Fusion SL Paper 2

Nuclear Fusion in Stars, Stellar Evolution, Stellar Parallax, Stellar Spectra

### Question-E.5 Fusion 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]

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

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]

## Markscheme

a.

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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