IB Physics HL Theme E: Nuclear and quantum physics questions

Exam-style multiple choice and written questions with full markschemes. Check multiple choice answers straight away, or type, photograph or draw a written answer and get it marked.

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Question 1

Multiple choiceHard1 mark

A hypothetical nuclear reaction combines nuclei with nucleon numbers 12 and 24 to form a nucleus with nucleon number 36. The combined initial rest mass is $M_i=35.70\,\mathrm{u}$ Use $1\,\mathrm{u}\,c^2=931.5\,\mathrm{MeV}$

Here, $\Delta m$ is the mass equivalent of the increase in total binding energy. What is $\Delta m/M_i$

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Answer: B. The graph gives binding energy per nucleon, so multiply each value by its nucleon number to obtain total binding energy. The initial total is $12(7.7)+24(8.0)=284.4\,\mathrm{MeV}$ and the final total is $36(8.5)=306.0\,\mathrm{MeV}$ Thus the increase is $21.6\,\mathrm{MeV}$ and

\frac{\Delta m}{M_i}=\frac{21.6}{(931.5)(35.70)}=6.50\times10^{-4}.

A: subtracts the per-nucleon values without weighting them by nucleon number, giving a negative result.
C: uses the initial total binding energy rather than its increase.
D: converts the energy increase to a mass in u but does not divide by the initial mass.
Remember: total binding energy is nucleon number multiplied by binding energy per nucleon.

Question 2

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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