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]
▶️Answer/Explanation
Markscheme
a.
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;
\({{\text{T}}_{\frac{1}{2}}} = \frac{{{\text{In}}2}}{{0.061}} = 11.4\left( {\text{s}} \right)\);
\(\left( {N = \frac{1}{8}{N_0}{\text{ so}}} \right)t = \left( {3{T_{\frac{1}{2}}} = } \right)34\left( {\text{s}} \right)\)
or
\(t = – \frac{{{\text{In}}0.125}}{{0.061}}\);
t=34(s);
This question is about nuclear physics and radioactive decay.
a.Define decay constant.[1]
\[{}_{86}^{210}{\rm{Rn}} \to {}_{84}^{206}{\rm{Po + X}}\]
(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]
Determine the distance of closest approach of particle X to the Au nucleus.[3]
▶️Answer/Explanation
Markscheme
a.
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);
\(0.5 = 1.6{e^{ – \lambda t}}\);
\(t = \left( { – \frac{{\ln \frac{{0.5}}{{1.6}}}}{\lambda } = } \right)\frac{{1.163}}{{8.0 \times {{10}^{ – 5}}}}\);
1.5×104(s);
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 \( = 3.7 \times {10^{ – 14}}\left( {\rm{m}} \right)\);