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Date May 2018 Marks available 3 Reference code 18M.2.SL.TZ1.6
Level Standard level Paper Paper 2 Time zone 1
Command term Show that 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]
a.

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

[2]
b.i.

After 4.3 × 106 years,

number of produced boron nuclei 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

4 10 Be 5 10 B + β + V ¯ e

conservation of mass number AND charge  5 10 B , 4 10 Be

 

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

[1 mark]

a.

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 = 1 8 , 12.5%, or 0.125

therefore 3 half lives have elapsed

t 1 2 = 4.3 × 10 6 3 = 1.43 × 10 6  «≈ 1.4 × 106» «y»

 

ALTERNATIVE 2

fraction of Be = 1 8 , 12.5%, or 0.125

1 8 = e λ ( 4.3 × 10 6 ) leading to λ = 4.836 × 10–7 «y»–1

ln 2 λ = 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.

λ = 2.90 × 10 3 253

= 1.1 × 10–5 «m»

 

Allow ECF from MP1 (incorrect temperature).

[2 marks]

c.iii.

correct units for Intensity (allow W, Nms1 OR Js1 in 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]
a.
[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.

Syllabus sections

Core » Topic 7: Atomic, nuclear and particle physics » 7.1 – Discrete energy and radioactivity
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