# IBDP Physics 12.2 – Nuclear physics: IB Style Question Bank HL Paper 2

### Question

The de Broglie wavelength λ of a particle accelerated close to the speed of light is approximately

λ = $\frac{hc}{E}$

where E is the energy of the particle.

A beam of electrons of energy 4.2 × 108 eV is produced in an accelerator.

(a) The electron beam is used to study the nuclear radius of carbon-12. The beam is directed from the left at a thin sample of carbon-12. A detector is placed at an angle θ  relative to the direction of the incident beam.

The graph shows the variation of the intensity of electrons with θ. There is a minimum of intensity for θ = θ0.

(i) Discuss how the results of the experiment provide evidence for matter waves.   [2]

(ii) The accepted value of the diameter of the carbon-12 nucleus is 4.94 × 1015 m.

Estimate the angle θ0 at which the minimum of the intensity is formed. [2]

(iii) Outline why electrons with energy of approximately 107 eV would be unsuitable for the investigation of nuclear radii.   [2]

(b) Experiments with many nuclides suggest that the radius of a nucleus is proportional to A , where A is the number of nucleons in the nucleus.

Show that the density of a nucleus remains approximately the same for all nuclei.  [2]

Ans

a.i

«the shape of the graph suggests that» electrons undergo diffraction «with carbon nuclei»

only waves diffract ✓

a.ii

$$sin \theta_0=\frac{2.96\times10^{-15}}{4.94\times10^{-15}}=0.599$$

37 «degrees» OR 0.64/0.65 «rad» ✓

a.iii

the de Broglie wavelength of electrons is «much» longer than the size of a nucleus ✓
hence electrons would not undergo diffraction
OR
no diffraction pattern would be observed ✓

b

volume of a nucleus proportional to $$\left ( A^{\frac{1}{3}} \right )^3$$ = A AND mass proportional to A ✓
the ratio $$\frac{mass}{volume}$$ independent of A «hence density the same for all nuclei» ✓

## Question

Rhodium-106 ($$_{\,\,\,45}^{106}{\text{Rh}}$$) decays into palladium-106 ($$_{\,\,\,46}^{106}{\text{Pd}}$$) by beta minus (β) decay. The diagram shows some of the nuclear energy levels of rhodium-106 and palladium-106. The arrow represents the β decay.

b.

Bohr modified the Rutherford model by introducing the condition mvr = n$$\frac{h}{{2\pi }}$$. Outline the reason for this modification.

[3]
c.i

Show that the speed v of an electron in the hydrogen atom is related to the radius r of the orbit by the expression

$v = \sqrt {\frac{{k{e^2}}}{{{m_{\text{e}}}r}}}$

where k is the Coulomb constant.

[1]
c.ii.

Using the answer in (b) and (c)(i), deduce that the radius r of the electron’s orbit in the ground state of hydrogen is given by the following expression.

$r = \frac{{{h^2}}}{{4{\pi ^2}k{m_{\text{e}}}{e^2}}}$

[2]
c.iiii.

Calculate the electron’s orbital radius in (c)(ii).

[1]
d.i..

Explain what may be deduced about the energy of the electron in the β decay.

[3]
d.ii.

Suggest why the β decay is followed by the emission of a gamma ray photon.

[1]
d.iii.

Calculate the wavelength of the gamma ray photon in (d)(ii).

[2]

## Markscheme

b.

the electrons accelerate and so radiate energy

they would therefore spiral into the nucleus/atoms would be unstable

electrons have discrete/only certain energy levels

the only orbits where electrons do not radiate are those that satisfy the Bohr condition «mvr = n$$\frac{h}{{2\pi }}$$»

[3 marks]

c.i

$$\frac{{{m_{\text{e}}}{v^2}}}{r} = \frac{{k{e^2}}}{{{r^2}}}$$

OR

KE = $$\frac{1}{2}$$PE hence $$\frac{1}{2}$$mev2 = $$\frac{1}{2}\frac{{k{e^2}}}{r}$$

«solving for v to get answer»

Answer given – look for correct working

[1 mark]

c.ii.

combining v = $$\sqrt {\frac{{k{e^2}}}{{{m_{\text{e}}}r}}}$$ with mevr = $$\frac{h}{{2\pi }}$$ using correct substitution

«eg $${m_e}^2\frac{{k{e^2}}}{{{m_{\text{e}}}r}}{r^2} = \frac{{{h^2}}}{{4{\pi ^2}}}$$»

correct algebraic manipulation to gain the answer

Answer given – look for correct working

Do not allow a bald statement of the answer for MP2. Some further working eg cancellation of m or r must be shown

[2 marks]

c.iiii.

« r = $$\frac{{{{(6.63 \times {{10}^{ – 34}})}^2}}}{{4{\pi ^2} \times 8.99 \times {{10}^9} \times 9.11 \times {{10}^{ – 31}} \times {{(1.6 \times {{10}^{ – 19}})}^2}}}$$»

r = 5.3 × 10–11 «m»

[1 mark]

d.i..

the energy released is 3.54 – 0.48 = 3.06 «MeV»

this is shared by the electron and the antineutrino

so the electron’s energy varies from 0 to 3.06 «MeV»

[3 marks]

d.ii.

the palladium nucleus emits the photon when it decays into the ground state «from the excited state»

[1 mark]

d.iii.

Photon energy

E = 0.48 × 106 × 1.6 × 10–19 = «7.68 × 10–14 J»

λ = «$$\frac{{hc}}{E} = \frac{{6.63 \times {{10}^{ – 34}} \times 3 \times {{10}^8}}}{{7.68 \times {{10}^{ – 14}}}}$$ =» 2.6 × 10–12 «m»

Award [2] for a bald correct answer

Allow ECF from incorrect energy

[2 marks]

## Question

The radioactive nuclide beryllium-10 (Be-10) undergoes beta minus (β–) decay to form a stable boron (B) nuclide.

The initial number of nuclei in a pure sample of beryllium-10 is N0. The graph shows how the number of remaining beryllium nuclei in the sample varies with time.

An ice sample is moved to a laboratory for analysis. The temperature of the sample is –20 °C.

a.

Identify the missing information for this decay.

[2]
b.iii.

Beryllium-10 is used to investigate ice samples from Antarctica. A sample of ice initially contains 7.6 × 1011 atoms of beryllium-10. The present activity of the sample is 8.0 × 10−3 Bq.

Determine, in years, the age of the sample.

[3]
c.iv.

The temperature in the laboratory is higher than the temperature of the ice sample. Describe one other energy transfer that occurs between the ice sample and the laboratory.

[2]

## Markscheme

a.

$$_{{\mkern 1mu} {\mkern 1mu} 4}^{10}{\text{Be}} \to _{{\mkern 1mu} {\mkern 1mu} 5}^{10}{\text{B}} + _{ – 1}^{\,\,\,0}{\text{e}} + {\overline {\text{V}} _{\text{e}}}$$

antineutrino AND charge AND mass number of electron $$_{ – 1}^{\,\,\,0}{\text{e}}$$, $$\overline {\text{V}}$$

conservation of mass number AND charge $$_{\,\,5}^{10}{\text{B}}$$, $$_{{\mkern 1mu} {\mkern 1mu} 4}^{10}{\text{Be}}$$

Do not accept V.

Accept $${\bar V}$$ without subscript e.

[2 marks]

b.iii.

λ «= $$\frac{{\ln 2}}{{1.4 \times {{10}^6}}}$$» = 4.95 × 10–7 «y–1»

rearranging of A = λN0eλt to give –λt = ln $$\frac{{8.0 \times {{10}^{-3}} \times 365 \times 24 \times 60 \times 60}}{{4.95 \times {{10}^{-7}} \times 7.6 \times {{10}^{11}}}}$$ «= –0.400»

t = $$\frac{{ – 0.400}}{{ – 4.95 \times {{10}^{ – 7}}}} = 8.1 \times {10^5}$$ «y»

Allow ECF from MP1

[3 marks]

c.iv.

from the laboratory to the sample

conduction – contact between ice and lab surface.

OR

convection – movement of air currents

Must clearly see direction of energy transfer for MP1.

Must see more than just words “conduction” or “convection” for MP2.

[2 marks]

[N/A]