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Date May 2018 Marks available 3 Reference code 18M.2.HL.TZ1.6
Level Higher 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.

[2]
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 nucleinumber of remaining beryllium nuclei=7.number of produced boron nucleinumber 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. The present activity of the sample is 8.0 × 10−3 Bq.

Determine, in years, the age of the sample.

[3]
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.

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.

[2]
c.iv.

Markscheme

104Be105B+01e+¯Ve104Be105B+01e+¯¯¯¯Ve

antineutrino AND charge AND mass number of electron 01e01e, ¯V¯¯¯¯V

conservation of mass number AND charge 105B105B, 104Be104Be

 

Do not accept V.

Accept ˉV¯V without subscript e.

[2 marks]

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 = 1818, 12.5%, or 0.125

therefore 3 half lives have elapsed

t12=4.3×1063=1.43×106t12=4.3×1063=1.43×106 «≈ 1.4 × 106» «y»

 

ALTERNATIVE 2

fraction of Be = 1818, 12.5%, or 0.125

18=eλ(4.3×106)18=eλ(4.3×106) leading to λ = 4.836 × 10–7 «y»–1

ln2λln2λ = 1.43 × 106 «y»

 

 

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

[3 marks]

b.ii.

λ «= ln21.4×106ln21.4×106» = 4.95 × 10–7 «y–1»

rearranging of AλN0eλt to give –λt = ln 8.0×103×365×24×60×604.95×107×7.6×10118.0×103×365×24×60×604.95×107×7.6×1011 «= –0.400»

t0.4004.95×107=8.1×1050.4004.95×107=8.1×105 «y»

 

 

Allow ECF from MP1

[3 marks]

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=2.90×103253λ=2.90×103253

= 1.1 × 10–5 «m»

 

Allow ECF from MP1 (incorrect temperature).

[2 marks]

c.iii.

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]

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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