Question

A radioactive element gets spilled over the floor of a room. Its half life period is 30 days. If the initial activity is ten times the permissible value, after how many days will it be safe to enter the room?

Answer 1

Use the relation between the half life and decay constant to determine the decay constant. It will be safe to enter the room if the final number of nuclei of radioactive elements is $\dfrac{1}{{10}}$ times the initial value. Then use the radioactive disintegration equation to determine the time.

Formula used:
$N\left( t \right) = N\left( 0 \right){e^{ - \lambda t}}$
Here, $N\left( t \right)$ is the number of nuclei of the radioactive element at time t, $N\left( 0 \right)$ is the original number of nuclei present in the radioactive element, $\lambda$ is the decay constant and t is the time.
The decay constant is given as,
$\lambda = \dfrac{{0.693}}{{{t_{1/2}}}}$
Here, ${t_{1/2}}$ is the half life of the radioactive element.

Complete step by step answer:
We have the law of radioactive disintegration,
$N\left( t \right) = N\left( 0 \right){e^{ - \lambda t}}$

Here, $N\left( t \right)$ is the number of nuclei of the radioactive element at time t, $N\left( 0 \right)$ is the original number of nuclei present in the radioactive element, $\lambda$ is the decay constant and t is the time.
We take the natural logarithm of the above equation as follows,
$\ln N\left( t \right) = \ln N\left( 0 \right) - \lambda t$
$\Rightarrow \lambda t = \ln N\left( 0 \right) - \ln N\left( t \right)$
$\Rightarrow t = \dfrac{1}{\lambda }\ln \left( {\dfrac{{N\left( 0 \right)}}{{N\left( t \right)}}} \right)$
We have given that the initial rate is ten times the permissible value. Therefore, we can say it will be safe to enter the room if the final number of nuclei of radioactive elements is $\dfrac{1}{{10}}$ times the initial value. Therefore, we can substitute $\dfrac{1}{{10}}N\left( 0 \right)$ for $N\left( t \right)$ in the above equation.
$\Rightarrow t = \dfrac{1}{\lambda }\ln \left( {\dfrac{{N\left( 0 \right)}}{{\dfrac{1}{{10}}N\left( 0 \right)}}} \right)$
$\Rightarrow t = \dfrac{1}{\lambda }\ln \left( {10} \right)$
$\Rightarrow t = \dfrac{{2.302}}{\lambda }$ …… (1)
Now, we have to determine the decay constant $\lambda$. We have given that the half life of the radioactive element is 30 days. Therefore, we use the relation between half life and decay constant as follows,
$\lambda = \dfrac{{0.693}}{{{t_{1/2}}}}$
Here, ${t_{1/2}}$ is the half life of the radioactive element.
We substitute 30 day for ${t_{1/2}}$ in the above equation.
$\lambda = \dfrac{{0.693}}{{30}}$
$\Rightarrow \lambda = 0.0231\,{\text{da}}{{\text{y}}^{ - 1}}$
We substitute $\lambda = 0.0231\,{\text{da}}{{\text{y}}^{ - 1}}$ in equation (1).
$t = \dfrac{{2.302}}{{0.0231\,{\text{da}}{{\text{y}}^{ - 1}}}}$
$\Rightarrow t = 100\,{\text{days}}$

Therefore, it will be safe to enter the room after 100 days.

Note:
To solve such types of questions on radioactivity, students should remember the formula for half life and radioactive decay equations of all kinds. After taking the natural logarithm, we have used the identity $\ln \left( {a - b} \right) = \ln \dfrac{a}{b}$ and $\ln \left( {{e^x}} \right) = x$. These formulae have crucial importance while solving the questions on radioactivity.