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 choiceMedium1 mark

The schematic shows two spherical nuclei. Assume that all nucleons have the same mass. Each option gives $R_X/R_Y$ and $\rho_X/\rho_Y$ respectively. Which row is correct?

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Answer: C. The nuclear radius relation is $R=R_0A^{1/3}$ From the figure, $R_X/R_Y=(8/64)^{1/3}=1/2$ The mass ratio is $1/8$ and the spherical volume ratio is $(1/2)^3=1/8$ so $\rho_X/\rho_Y=(1/8)/(1/8)=1$

A: uses the mass ratio as the density ratio, omitting the volume comparison.
B: reverses the radius ratio, although its density ratio is unity.
D: treats the masses as equal and uses the inverse volume ratio, giving $8$
Remember: nuclear mass and volume both scale with nucleon number, so nuclear density is constant.

Question 2

Multiple choiceMedium1 mark

Two non-relativistic particles, X and Y, have masses $m$ and $2m$ and speeds $v$ and $3v$ respectively. Which row gives the momentum ratio $p_Y/p_X$ and de Broglie wavelength ratio $\lambda_Y/\lambda_X$

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Answer: A. For non-relativistic particles, $p=mv$ so $p_Y/p_X=(2m)(3v)/(mv)=6$ Since $\lambda=h/p$ $\lambda_Y/\lambda_X=p_X/p_Y=1/6$

B: treats wavelength as directly proportional to momentum. C: omits the mass difference when comparing momenta, giving $p_Y/p_X=3$ D: reverses the ratios, giving $p_X/p_Y$ and $\lambda_X/\lambda_Y$ instead.

Remember: de Broglie wavelength is inversely proportional to momentum.

Question 3

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 4

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?

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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.

Question 5

Multiple choiceMedium1 mark

A fusion event in a star is represented below. The labels give the total rest masses before and after fusion. What fraction of the initial rest energy is released?

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Answer: B. The energy released is the decrease in rest energy, $E_{\rm rel}=(M_i-M_f)c^2$ The initial rest energy is $E_i=M_i c^2$ so $\dfrac{E_{\rm rel}}{E_i}=\dfrac{(M_i-M_f)c^2}{M_i c^2}=\dfrac{M_i-M_f}{M_i}$

A: divides the mass defect by the final mass, not the initial mass.
C: gives the fraction of the initial rest energy that remains as product rest energy.
D: takes the reciprocal of the required fraction.
Remember: the energy released in fusion comes from the decrease in rest energy.

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