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Date May 2012 Marks available 7 Reference code 12M.2.SL.TZ1.5
Level Standard level Paper Paper 2 Time zone Time zone 1
Command term Define, Determine, Explain, and State Question number 5 Adapted from N/A

Question

Part 2 Radioactive decay

Describe the phenomenon of natural radioactive decay.

[3]
a.

A nucleus of americium-241 (Am-241) decays into a nucleus of neptunium-237 (Np-237) in the following reaction.

\[{}_{95}^{241}{\rm{Am}} \to {}_X^{237}{\rm{Np}} + {}_2^4\alpha \]

(i) State the value of X.

(ii) Explain in terms of mass why energy is released in the reaction in (b).

(iii) Define binding energy of a nucleus.

(iv) The following data are available.

Determine the energy released in the reaction in (b).

[7]
b.

Markscheme

emission of (alpha/beta/gamma) particles/photons/electromagnetic radiation;
nucleus becomes more (energetically) stable;
constant probability of decay (per unit time);
is random process;
activity/number of unstable nuclei in sample reduces by half over constant time intervals/exponentially;
not affected by temperature/environment / is spontaneous process;

a.

(i) 93;

(ii) mass of products is less than mass of reactants / there is a mass defect;
mass is converted into energy (according to equation E=mc2);

(iii) the (minimum) energy required to (completely) separate the nucleons in a nucleus / the energy released when a nucleus is assembled from its constituent nucleons;

(iv) calculation of binding energies as shown below;

americium-241 = 241×7.54=1817.14 MeV
neptunium-237 = 237×7.58 = 1796.46 MeV
helium-4 = 4×7.07 = 28.28 MeV

energy released is the difference of binding energies;
and so equals 7.60 MeV;

Award [2 max] for an answer that multiplies by the number of neutrons or number of protons.
Ignore any negative sign in answer.

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

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