Question
(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 ) \(\begin{aligned} & \lambda=\ll \frac{\ln 2}{12.3} \rightsquigarrow 0.056 \kappa y^{-1} » \text { OR } 0.5^{\frac{1}{m}} \text { OR } e^{-1 \times \frac{\ln 2}{12.3}} \\ & 0.945 \text { OR } 94.5 \%\end{aligned}\)
Part 2 Radioactive decay
a.
Describe the phenomenon of natural radioactive decay.[3]
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;
(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.