Question

A Geiger counter registers a count rate of 8,000 counts per minute from a sample of a radioisotope. The count rate 24 minutes later is 1,000 counts per minute. What is the half-life of the radioisotope?

Answer 1

11.54 minutes

Explanation:

The decay rate equation is given by

$N = N_0e^{-\frac{t}{\lambda}}$

where $\lambda$ is the half-life. We can rewrite this as

$\dfrac{N}{N_0} = e^{-\frac{t}{\lambda}}$

Taking the natural logarithm of both sides, we get

$\ln \left(\dfrac{N}{N_0}\right) = -\left(\dfrac{t}{\lambda}\right)$

Solving for $\lambda$,

$\lambda = -\dfrac{t}{\ln \left(\frac{N}{N_0}\right)}$

$\:\:\:\:= -\dfrac{(24\:\text{minutes})}{\ln \left(\frac{1000\:\text{counts/min}}{8000\:\text{counts/min}}\right)}$

$\:\:\:\:=11.54\:\text{minutes}$