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Date	May 2018	Marks available	3	Reference code	18M.2.HL.TZ2.9
Level	Higher level	Paper	Paper 2	Time zone	2
Command term	Explain	Question number	9	Adapted from	N/A

Question

Rhodium-106 ( $_{45}^{106} Rh$ $_{\,\,\,45}^{106}{\text{Rh}}$ ) decays into palladium-106 ( $_{46}^{106} Pd$ $_{\,\,\,46}^{106}{\text{Pd}}$ ) 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.

M18/4/PHYSI/HP2/ENG/TZ2/09.d

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

[3]

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

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

where k is the Coulomb constant.

[1]

c.i.

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.

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

[2]

c.ii.

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

[1]

c.iii.

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

[3]

d.i.

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

[1]

d.ii.

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

[2]

d.iii.

Markscheme

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 $\frac{h}{2 π}$ $\frac{h}{{2\pi }}$ »

[3 marks]

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

KE = $\frac{1}{2}$ $\frac{1}{2}$ PE hence $\frac{1}{2}$ $\frac{1}{2}$ m_ev² = $\frac{1}{2} \frac{k e^{2}}{r}$ $\frac{1}{2}\frac{{k{e^2}}}{r}$

«solving for v to get answer»

Answer given – look for correct working

[1 mark]

c.i.

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

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

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.ii.

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

r = 5.3 × 10^–11 «m»

[1 mark]

c.iii.

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.i.

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

[1 mark]

d.ii.

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]

d.iii.

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

d.iii.

Syllabus sections

Additional higher level (AHL) » Topic 12: Quantum and nuclear physics » 12.2 – Nuclear physics

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