Question

A count rate meter is used to measure the activity of a given sample. At one instant, the meter shows 4750 counts per minute. Five minutes later, it shows 2700 counts per minute. Find (a) the decay constant and (b) the half-life of the sample. $(lo{g}_{10}1.760=0.2455)$

Answer 1

We have: $N=N_0{e}^{-\lambda t}$ (i)
The activity of the sample is given by: $\frac{dN}{dt}=-\lambda N_o{e}^{-\lambda t}=-\lambda N$
i.e., activity is proportional to the number of undecayed nuclei.

At $t=0,{\left(\frac{dN}{dt}\right)}_{t=0}=-\lambda N$

At $t=5min,{\left(\frac{dN}{dt}\right)}_{t=5min}=-\lambda N$

$\therefore \frac{N_0}{N}=\frac{{\left(\frac{dN}{dt}\right)}_{t=0}}{{\left(\frac{dN}{dt}\right)}_{t=5min}}=\frac{4750}{2700}=1.760$

a) From Eq. (i), we have: $\frac{N}{N_0}=e^{-\lambda t}$
$\Rightarrow \frac{N_0}{N}=e^{\lambda t}$

$\Rightarrow lo{g}_e\frac{N_0}{N}=\lambda t$

$\Rightarrow \lambda =\frac{1}{t}lo{g}_e\frac{N_0}{N}=\frac{2.3026}{t}\times lo{g}_{10}\left(\frac{N_0}{N}\right)$ (iii)

Substituting the given values, we have:
$\lambda =\frac{2.3026}{5}\times 0.2455=0.113mi{n}^{-1}$

b) Half life of sample, $T=\frac{0.6931}{\lambda }=\frac{0.6931}{0.113}=6.1mi{n}^{-1}$