CBSE 2022 · Region 5 · Set 3 · Q8 · 3 marks
In a fission event of $\displaystyle { }_{92}^{238} \mathrm{U}$ by fast moving neutrons, no neutrons are emitted and final products, after the beta decay of the primary fragments, are $\displaystyle { }_{58}^{140} \mathrm{Ce}$ and $\displaystyle { }_{44}^{99} \mathrm{Ru}$. Calculate Q for this process. Neglect the masses of electrons/ positrons emitted during the intermediate steps. Given : $\displaystyle \mathrm{m}\left({ }_{92}^{238} \mathrm{U}\right)=238.05079 \mathrm{u} ; \mathrm{m}\left({ }_{58}^{140} \mathrm{Ce}\right)=139.90543 \mathrm{u}$ \[\mathrm{m}\left({ }_{44}^{99} \mathrm{Ru}\right)=98.90594 \mathrm{u} ; \mathrm{m}\binom{1}{0}=1.008665 \mathrm{u} \]
Marking-scheme solution
$\displaystyle \Delta \mathrm{m}=$ total mass of the reactants - total mass of the products=\left[m\left({ }_{92}^{238} \mathrm{U}\right)+m_{n}-m\left({ }_{58}^{140} \mathrm{Ce}\right)-m\left({ }_{44}^{99} \mathrm{Ru}\right)\right]$\displaystyle =[238 \cdot 05079+1 \cdot 008665-139 \cdot 90543-98 \cdot 90594] u$
$\displaystyle =[239 \cdot 059455-238 \cdot 81137] u$=0 \cdot 248085^{u}Q -value $\displaystyle =0 \cdot 248085 \times 931.5 \mathrm{MeV}$=231.09 \mathrm{MeV}
$$
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CBSE Class 12 Physics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.