# 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 =

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.   

(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. 

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

(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.  

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.


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.


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.

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


c.iiii.

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


d.i..

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


d.ii.

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


d.iii.

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



## 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  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. 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.


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.



## 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]