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Question 18M.2.HL.TZ2.9d.i

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Date May 2018 Marks available [Maximum mark: 3] Reference code 18M.2.HL.TZ2.9d.i
Level HL Paper 2 Time zone TZ2
Command term Explain Question number d.i Adapted from N/A
d.i.
[Maximum mark: 3]
18M.2.HL.TZ2.9d.i

Rhodium-106 ( 45 106 Rh ) decays into palladium-106 ( 46 106 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

(d.i)

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

[3]

Markscheme

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]

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