IB Physics HL E.4 practice question

Question 1

Multiple choiceHard1 mark

A neutron initiates the reaction

{}^{1}_{0}\mathrm n+{}^{235}_{92}\mathrm U \rightarrow {}^{141}_{56}\mathrm{Ba}+{}^{92}_{36}\mathrm{Kr}+3\,{}^{1}_{0}\mathrm n .

Each fission releases three neutrons, and on average a fraction $r$ of neutrons from one generation cause a fission in the next, where $0<3r<1$ Let $m_{235}$ $m_{141}$ $m_{92}$ and $m_n$ be the nuclear masses of the indicated species. Which expression gives the expected total energy released by the complete chain?

Show answer and explanation

Answer: A. The mass defect per fission is $(m_{235}+m_n)-(m_{141}+m_{92}+3m_n)=m_{235}-m_{141}-m_{92}-2m_n$ so one fission releases $Q=\Delta mc^2$ Each fission produces an average of $3r$ further fissions. If $N$ is the expected total number of fissions, $N=1+3rN$ giving $N=1/(1-3r)$ The total expected energy is $NQ$

B: omits the net mass of the two additional neutrons in the products.
C: uses $r$ rather than $3r$ as the average number of further fissions per fission.
D: counts the initial fission and only its immediate descendants, omitting later generations.
Remember: the chain multiplication factor is the neutron yield per fission multiplied by the fraction that cause further fissions.

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