Question

An experiment requires minimum beta activity produced at the rate of 346 beta particles per minute. The half-life period of 42 mo99 which is a beta emitter is 66.6 h. Find the minimum amount of 42 mo99 required to carry out the experiment in 6.909 h.

Answer 1

This question requires that we use the equation for finding the amount of radioactive atoms remaining after a certain period of time. In this question, we require that in the end , there are at least 346 atoms remaining after a period of 6.909h. The half life for MO99 is 66.6h.

The formula for the amount of substance remaining a time $t$ is,

$N(t)=N_0e^{-\tfrac{0.693t}{t_{{1}/{2}}}$,

where $N_0$ is the initial amount of the radioactive source ,  $t_{{1}/{2}}$ is the half life,  $N(t)$  amount present at any time,and  $t$ is the time since the beginning of the experiment.

We can then solve for the initial amount $N_0$, so that

$N_0=\frac{N(t)}{e^{-\tfrac{0.693t}{t_{{1}/{2}}}}}=\frac{346}{e^{-\tfrac{0.693\times 6.909}{66.6}}}}=371.79$

There has to be 372 atoms of M099 in the beginning of the experiment if it .