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.

Question 1

Multiple choiceMedium1 mark

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

Nucleus Xnucleon number A=8RXNucleus Ynucleon number A=64RY
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Answer: C. The nuclear radius relation is R=R0A1/3R=R_0A^{1/3} From the figure, RX/RY=(8/64)1/3=1/2R_X/R_Y=(8/64)^{1/3}=1/2 The mass ratio is 1/81/8 and the spherical volume ratio is (1/2)3=1/8(1/2)^3=1/8 so ρX/ρY=(1/8)/(1/8)=1\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 88
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 mm and 2m2m and speeds vv and 3v3v respectively. Which row gives the momentum ratio pY/pXp_Y/p_X and de Broglie wavelength ratio λY/λX\lambda_Y/\lambda_X

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Answer: A. For non-relativistic particles, p=mvp=mv so pY/pX=(2m)(3v)/(mv)=6p_Y/p_X=(2m)(3v)/(mv)=6 Since λ=h/p\lambda=h/p λY/λX=pX/pY=1/6\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 pY/pX=3p_Y/p_X=3 D: reverses the ratios, giving pX/pYp_X/p_Y and λX/λY\lambda_X/\lambda_Y instead.

Remember: de Broglie wavelength is inversely proportional to momentum.

Question 3

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?

before fusionreactantstotal rest mass Miafter fusionproductstotal rest mass Mffusionreleased energyMi > Mf
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Answer: B. The energy released is the decrease in rest energy, Erel=(Mi−Mf)c2E_{\rm rel}=(M_i-M_f)c^2 The initial rest energy is Ei=Mic2E_i=M_i c^2 so ErelEi=(Mi−Mf)c2Mic2=Mi−MfMi\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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