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

Paper 2Hard10 marks

A gas is studied using an emission spectrum and an absorption spectrum. For the absorption measurement, the atoms are initially in their lowest energy state. Figure 1 shows the resolved line positions; each line-centre wavelength has an uncertainty of ±2 nm\pm 2\ \text{nm}

Assume each line corresponds to a photon transition. Take the lowest atomic energy to be 0 eV0\ \text{eV} and use hc=1240 eV nmhc=1240\ \text{eV nm}

Relative intensityEmission00.51415 nm620 nm1240 nmRelative intensityAbsorption00.51415 nm1240 nm3504004505005506006507007508008509009501000105011001150120012501300Wavelength / nm
(a)

State the feature of the emission spectrum that provides evidence for discrete atomic energy levels.

[1]
(b)

Calculate the energy of a photon corresponding to the emission line at 415 nm415\ \text{nm}

[2]
(c)

Calculate the energy gap corresponding to the absorption line at 1240 nm1240\ \text{nm}

[1]
(d)

Using the results from (b) and (c) and the information in Figure 1, determine the minimum number of energy levels needed to account for all the displayed lines. Draw an energy-level diagram, with levels labelled relative to the lowest energy state, and show the transitions corresponding to every displayed emission and absorption line.

[4]
(e)

Explain why the absence of a 620 nm620\ \text{nm} absorption line does not contradict the 620 nm620\ \text{nm} emission line. State the initial state required for absorption of a 620 nm620\ \text{nm} photon.

[2]

Question 2

Paper 2Hard10 marks

A monochromatic X-ray photon scatters from an electron that is initially at rest. Model the interaction as a single photon–electron event, with no energy transferred to the rest of the material. Treat the recoiling electron non-relativistically.

Figure 1 shows the geometry and the measured photon wavelengths. The wavelengths are rounded to the nearest 0.00001 nm0.00001\ \mathrm{nm}

xyincident photon, λi=0.08000 nmscattered photon, λs=0.08243 nm90°Oelectron initially at rest

Use h=6.626×10−34 J sh=6.626\times10^{-34}\ \mathrm{J\,s} c=2.998×108 m s−1c=2.998\times10^8\ \mathrm{m\,s^{-1}} and me=9.11×10−31 kgm_e=9.11\times10^{-31}\ \mathrm{kg}

(a)

Calculate the increase in wavelength of the photon.

[1]
(b)

Calculate the kinetic energy transferred to the electron, in J.

[2]
(c)

Determine the speed and direction of the recoiling electron. Give the direction relative to the positive xx-axis.

[4]
(d)

Calculate the electron’s kinetic energy using your result from (c). Comment on whether this is consistent with your answer to (b), taking the precision of the wavelength measurements into account.

[3]

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 Mi=35.70 uM_i=35.70\,\mathrm{u} Use 1 u c2=931.5 MeV1\,\mathrm{u}\,c^2=931.5\,\mathrm{MeV}

040801201602002400246810nucleon number, Abinding energy per nucleon / MeV(1, 0.0)(12, 7.7)(24, 8.0)(36, 8.5)(56, 8.8)(100, 8.4)(240, 7.6)

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

Show answer and explanation

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 MeV12(7.7)+24(8.0)=284.4\,\mathrm{MeV} and the final total is 36(8.5)=306.0 MeV36(8.5)=306.0\,\mathrm{MeV} Thus the increase is 21.6 MeV21.6\,\mathrm{MeV} and

ΔmMi=21.6(931.5)(35.70)=6.50×10−4.\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

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?

Show answer and explanation

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.

Question 5

Paper 2Hard10 marks

Figure 1 shows two fission channels with the same products. The masses shown are atomic masses; the electron numbers balance in each reaction. The incident neutron’s kinetic energy is 0.025 eV0.025\ \text{eV} Use 1 u c2=931.5 MeV1\ \text{u}\,c^2=931.5\ \text{MeV}

spontaneous fission23692U14054Xe+9438Sr+10n+10n236.045568 u139.921640 u93.915356 u1.008665 u1.008665 uneutron-induced fission10n+23592U14054Xe+9438Sr+10n+10n1.008665 u235.043930 u139.921640 u93.915356 u1.008665 u1.008665 ukinetic energy0.025 eV
(a)

Calculate the energy released in the spontaneous fission reaction shown in Figure 1, in MeV.

[3]
(b)

Calculate the energy released in the neutron-induced fission reaction shown in Figure 1, in MeV.

[3]
(c)

A student claims that the greater energy release calculated for neutron-induced fission is supplied by the incident neutron’s kinetic energy. Assess this claim using Figure 1 and your answers. Include a quantitative comparison and account for the difference between the two energy releases.

[4]

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