Nuclear Physics: Question 3

Worked solution

Part (a): Show that the half-life is about 12.5 hours

The decay constant and half-life are related by: $t_{1/2}=\frac{\ln2}{\lambda}$

Substituting the given value: $t_{1/2}=\frac{0.6931}{1.54\times10^{-5}}=4.501\times10^4\text{ s}$

Converting to hours (dividing by $3600\text{ s h}^{-1}$): $t_{1/2}=\frac{4.501\times10^4}{3600}=12.50\text{ h}$

Recompute as a check, working backwards: if $t_{1/2}=12.50\text{ h}=4.500\times10^4\text{ s}$, then $\lambda=\ln2/t_{1/2}=0.6931/(4.500\times10^4)=1.54\times10^{-5}\text{ s}^{-1}$, matching the given decay constant.

So $t_{1/2}\approx\boxed{12.5}\text{ hours}$, as required to show.

Part (b): Initial number of undecayed nuclei

Activity and the number of undecayed nuclei are related by $A=\lambda N$, so at $t=0$: $N_0=\frac{A_0}{\lambda}=\frac{8.00\times10^{10}}{1.54\times10^{-5}}$

$N_0=5.195\times10^{15}$

Recompute as a check using a different split of the powers of ten: $\dfrac{8.00}{1.54}=5.195$, and $\dfrac{10^{10}}{10^{-5}}=10^{15}$, giving the same $N_0=5.195\times10^{15}$.

$N_0=\boxed{5.19\times10^{15}}\text{ nuclei (3 s.f.)}$

Part (c): Activity after 24.0 hours

First convert the elapsed time to seconds, since $\lambda$ is in $\text{s}^{-1}$: $t=24.0\text{ h}\times3600\text{ s h}^{-1}=8.64\times10^4\text{ s}$

Activity obeys the same exponential decay law as the number of nuclei: $A=A_0e^{-\lambda t}$

The exponent is: $\lambda t=1.54\times10^{-5}\times8.64\times10^4=1.331$

So: $A=8.00\times10^{10}\times e^{-1.331}=8.00\times10^{10}\times0.2643$

$A=2.115\times10^{10}\text{ Bq}$

Recompute as a check using the half-life instead: $24.0\text{ h}$ is $24.0/12.50=1.920$ half-lives, so $A=A_0\times(0.5)^{1.920}=8.00\times10^{10}\times0.2644=2.115\times10^{10}\text{ Bq}$, the two methods agree.

$A=\boxed{2.11\times10^{10}}\text{ Bq (3 s.f.)}$

Final answers

• (a) $t_{1/2}\approx\boxed{12.5}\text{ hours}$
• (b) $N_0=\boxed{5.19\times10^{15}}$ nuclei
• (c) $A=\boxed{2.11\times10^{10}}\text{ Bq}$