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Date	May 2018	Marks available	2	Reference code	18M.2.SL.TZ1.6
Level	Standard level	Paper	Paper 2	Time zone	1
Command term	Derive	Question number	6	Adapted from	N/A

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.

Identify the missing information for this decay.

[1]

On the graph, sketch how the number of boron nuclei in the sample varies with time.

[2]

b.i.

After 4.3 × 10⁶ years,

$\frac{{{\text{number of produced boron nuclei}}}}{{{\text{number of remaining beryllium nuclei}}}} = 7.$

Show that the half-life of beryllium-10 is 1.4 × 10⁶ years.

[3]

b.ii.

Beryllium-10 is used to investigate ice samples from Antarctica. A sample of ice initially contains 7.6 × 10¹¹ atoms of beryllium-10. State the number of remaining beryllium-10 nuclei in the sample after 2.8 × 10⁶ years.

[1]

b.iii.

State what is meant by thermal radiation.

[1]

c.i.

Discuss how the frequency of the radiation emitted by a black body can be used to estimate the temperature of the body.

[2]

c.ii.

Calculate the peak wavelength in the intensity of the radiation emitted by the ice sample.

[2]

c.iii.

Derive the units of intensity in terms of fundamental SI units.

[2]

c.iv.

Markscheme

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

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

Correct identification of both missing values required for [1].

[1 mark]

correct shape ie increasing from 0 to about 0.80 N₀

crosses given line at 0.50 N₀

M18/4/PHYSI/SP2/ENG/TZ1/06.b.i/M

[2 marks]

b.i.

ALTERNATIVE 1

fraction of Be = $\frac{1}{8}$ , 12.5%, or 0.125

therefore 3 half lives have elapsed

${t_{\frac{1}{2}}} = \frac{{4.3 \times {{10}^6}}}{3} = 1.43 \times {10^6}$ «≈ 1.4 × 10⁶» «y»

ALTERNATIVE 2

fraction of Be = $\frac{1}{8}$ , 12.5%, or 0.125

$\frac{1}{8} = {{\text{e}}^{ - \lambda }}(4.3 \times {10^6})$ leading to λ = 4.836 × 10^–7 «y»^–1

$\frac{{\ln 2}}{\lambda }$ = 1.43 × 10⁶ «y»

Must see at least one extra sig fig in final answer.

[3 marks]

b.ii.

1.9 × 10¹¹

[1 mark]

b.iii.

emission of (infrared) electromagnetic/infrared energy/waves/radiation.

[1 mark]

c.i.

the (peak) wavelength of emitted em waves depends on temperature of emitter/reference to Wein’s Law

so frequency/color depends on temperature

[2 marks]

c.ii.

$\lambda = \frac{{2.90 \times {{10}^{ - 3}}}}{{253}}$

= 1.1 × 10^–5 «m»

Allow ECF from MP1 (incorrect temperature).

[2 marks]

c.iii.

correct units for Intensity (allow W, Nms^–¹OR Js^–¹in numerator)

rearrangement into proper SI units = kgs^–3

Allow ECF for MP2 if final answer is in fundamental units.

[2 marks]

c.iv.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

[N/A]

c.iv.

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Core » Topic 8: Energy production » 8.1 – Energy sources

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