Define the decay constant of a radioactive isotope.

Show that the decay constant \(\lambda \) is related to the half-life \({T_{\frac{1}{2}}}\) by the expression

\[\lambda {T_{\frac{1}{2}}} = \ln 2.\]

Strontium-90 is a radioactive isotope with a half-life of 28 years. Calculate the time taken for 65% of the strontium-90 nuclei in a sample of the isotope to decay.

probability of decay (of a nucleus) per unit time;

Accept \(\frac{A}{N}\) with symbols defined.

\(N = {N_0}{{\text{e}}^{ - \lambda t}}\) and \(N = \frac{{{N_0}}}{2}\) when \(t = {T_{\frac{1}{2}}}\)\(\,\,\,\)or\(\,\,\,\)\(\frac{{{N_0}}}{2} = {N_0}{{\text{e}}^{ - \lambda {T_{\frac{1}{2}}}}}\);

\(\frac{1}{2} = {{\text{e}}^{ - \lambda {T_{\frac{1}{2}}}}}\)\(\,\,\,\)or\(\,\,\,\)\(2 = {{\text{e}}^{\lambda {T_{\frac{1}{2}}}}}\);

\(\left( {{\text{so }}\ln 2 = \lambda {T_{\frac{1}{2}}}} \right)\)

Answer given, award marks for correct working only.

\(\lambda  = \frac{{\ln 2}}{{28}}\)\(\,\,\,\)or\(\,\,\,\)\(0.025\left( {{{\text{y}}^{ - 1}}} \right]\);

\(0.35 = {{\text{e}}^{ - 0.025t}}\);

\(t = 42{\text{ (years)}}\);

Award [3] for a bald correct answer.

Award [2 max] for an answer of 17 years (using 0.65 instead of 0.35).

or

\(0.35 = {\left[ {\frac{1}{2}} \right]^x}\) where \(x = \frac{t}{{{T_{\frac{1}{2}}}}}\);

\(\frac{t}{{{T_{\frac{1}{2}}}}} = 1.5\);

\(t = 42{\text{ (years)}}\);

Award [3] for a bald correct answer.

Award [2 max] for an answer of 17 years (using 0.65 instead of 0.35).

(a) was an easy mark.

In (b) about half of the candidates could derive the relationship between half-life and decay constant, but many were completely lost.

The half-life calculation in (c) was generally well done, but a common mistake was to use 0.65 as the fraction remaining.