*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 × 10^{8 }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 = θ_{0}.

(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 × 10^{–15} m.

Estimate the angle θ_{0} at which the minimum of the intensity is formed. [2]

(iii) Outline why electrons with energy of approximately 10^{7} 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]

**Answer/Explanation**

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.

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.

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.

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

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

Explain what may be deduced about the energy of the electron in the *β*^{–} decay.

Suggest why the *β*^{–} decay is followed by the emission of a gamma ray photon.

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

**Answer/Explanation**

## 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]**

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

*OR*

KE = \(\frac{1}{2}\)PE hence \(\frac{1}{2}\)*m*_{e}*v*^{2} = \(\frac{1}{2}\frac{{k{e^2}}}{r}\)

**«**solving for *v *to get answer**»**

*Answer given – look for correct working*

*[1 mark]*

combining *v* = \(\sqrt {\frac{{k{e^2}}}{{{m_{\text{e}}}r}}} \) with *m*_{e}*vr* = \(\frac{h}{{2\pi }}\) 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]*

**«** *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]**

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]**

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

**[1 mark]**

Photon energy

*E* = 0.48 × 10^{6} × 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 N_{0}. 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.

Beryllium-10 is used to investigate ice samples from Antarctica. A sample of ice initially contains 7.6 × 10^{11} atoms of beryllium-10. The present activity of the sample is 8.0 × 10^{−3} Bq.

Determine, in years, the age of the sample.

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.

**Answer/Explanation**

## 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

*mass number of electron \(_{ – 1}^{\,\,\,0}{\text{e}}\), \(\overline {\text{V}} \)*

**AND**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]**

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

rearranging of *A* = *λN*_{0}e^{–λt} 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]*

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]