IB DP Physics 12.2 – Nuclear physics Question Bank SL Paper 3


This question is about radioactive decay.

A nucleus of magnesium-23 decays forming a nucleus of sodium-23 with the emission of an electron neutrino and a β+ particle.

a.Outline why the existence of neutrinos was hypothesized to account for the energy spectrum of beta decay.[3]

b.The decay constant for magnesium-23 is 0.061 s−1. Calculate the time taken for the number of magnesium-23 nuclei to fall to 12.5% of its initial value.[2]



spectrum of beta decay is continuous;
with a maximum value of energy;
the resulting energy difference between energy of any
β(+) and maximum β(+) energy is accounted for by the energy of the neutrino / reference to energy difference between parent energy level and excited energy level of daughter;


\({{\text{T}}_{\frac{1}{2}}} = \frac{{{\text{In}}2}}{{0.061}} = 11.4\left( {\text{s}} \right)\);
\(\left( {N = \frac{1}{8}{N_0}{\text{ so}}} \right)t = \left( {3{T_{\frac{1}{2}}} = } \right)34\left( {\text{s}} \right)\)


\(t = – \frac{{{\text{In}}0.125}}{{0.061}}\);


This question is about nuclear physics and radioactive decay.

a.Define decay constant.[1]

b.A sample of 1.6 mol of the radioactive nuclide radon-210 \(\left( {{}_{86}^{210}{\rm{Rn}}} \right)\) decays into polonium-206 \(\left( {{}_{84}^{206}{\rm{Po}}} \right)\) with the production of one other particle.

\[{}_{86}^{210}{\rm{Rn}} \to {}_{84}^{206}{\rm{Po + X}}\]

(i) Identify particle X.
(ii) The radioactive decay constant of radon-210 is 8.0×10–5s–1. Determine the time required to produce 1.1 mol of polonium-206.[5]

c.Particle X has an initial kinetic energy of 6.2MeV after the decay in (b). In a scattering experiment, particle X is aimed head-on at a stationary gold-197 \(\left( {{}_{76}^{197}{\rm{Au}}} \right)\) nucleus.

Determine the distance of closest approach of particle X to the Au nucleus.[3]




the probability of decay of a nucleus per unit time;


(i) alpha particle / helium nucleus;

(ii) number of Po nuclei produced=number of Rn nuclei decayed (seen or implied);
\(0.5 = 1.6{e^{ – \lambda t}}\);
\(t = \left( { – \frac{{\ln \frac{{0.5}}{{1.6}}}}{\lambda } = } \right)\frac{{1.163}}{{8.0 \times {{10}^{ – 5}}}}\);


initial kinetic energy=electric potential energy at closest distance;
kinetic energy \(E = \left( {6.2 \times {{10}^6} \times 1.6 \times {{10}^{ – 19}} = } \right)9.9 \times {10^{ – 13}}\left( {\rm{J}} \right)\);
\(d = k\frac{{{q_1}{q_2}}}{E} = 8.99 \times {10^9}\frac{{2 \times 79 \times {{\left[ {1.6 \times {{10}^{ – 19}}} \right]}^2}}}{{9.9 \times {{10}^{ – 13}}}}\left( {\rm{m}} \right)\) or \( = 3.7 \times {10^{ – 14}}\left( {\rm{m}} \right)\);

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