IB Physics HL E.4 Fission questions

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Question 1

Multiple choiceHard1 mark

A neutron initiates the reaction

01n+92235U→56141Ba+3692Kr+3 01n.{}^{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 rr of neutrons from one generation cause a fission in the next, where 0<3r<10<3r<1 Let m235m_{235} m141m_{141} m92m_{92} and mnm_n be the nuclear masses of the indicated species. Which expression gives the expected total energy released by the complete chain?

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Answer: A. The mass defect per fission is (m235+mn)−(m141+m92+3mn)=m235−m141−m92−2mn(m_{235}+m_n)-(m_{141}+m_{92}+3m_n)=m_{235}-m_{141}-m_{92}-2m_n so one fission releases Q=Δmc2Q=\Delta mc^2 Each fission produces an average of 3r3r further fissions. If NN is the expected total number of fissions, N=1+3rNN=1+3rN giving N=1/(1−3r)N=1/(1-3r) The total expected energy is NQNQ

B: omits the net mass of the two additional neutrons in the products.
C: uses rr rather than 3r3r 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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