Question

An open container has a net 10-gram mass of a radioactive material. The net mass in the container after 2 mean lives is approximately?
(A) 1.35 gram
(B) 2.5 gram
(C) 10 gram
(D) 5 gram

Answer 1

Use the exponential form of the decay equation to determine the net mass of the material. The mean-life of the sample is inverse of the decay constant.

Formula used:
$N = {N_0}{e^{ - \lambda t}}$
Here, ${N_0}$ is the actual number of particles in the material at $t = 0$, N is the number of particles remaining in the material at time t and $\lambda$ is the decay constant.

Complete step by step answer:
A radioactive material undergoes exponential decay following the decay equation,
$\Rightarrow N = {N_0}{e^{ - \lambda t}}$
Here, ${N_0}$ is the actual number of particles in the material at $t = 0$, N is the number of particles remaining in the material at time t and $\lambda$ is the decay constant.
The mean life of the radioactive sample is defined as the average lifetime of the sample before decay. It is denoted by $\tau$.
The mean life of the radioactive sample is the inverse of the decay constant. Therefore,
$\Rightarrow \tau = \dfrac{1}{\lambda }$
We were asked to determine the mass of the material after two mean lives. That is $t = 2\tau$.
Therefore,
$\Rightarrow t = 2\left( {\dfrac{1}{\lambda }} \right) = \dfrac{2}{\lambda }$
Substitute 10 gram for ${N_0}$ and $\dfrac{2}{\lambda }$ for t in equation (1).
$\Rightarrow N = \left( {10\,gram} \right){e^{ - \lambda \left( {\dfrac{2}{\lambda }} \right)}}$
$\Rightarrow N = \left( {10\,gram} \right){e^{ - 2}}$
$\Rightarrow N = \left( {10\,gram} \right)\left( {0.135} \right)$
$\Rightarrow \therefore N = 1.35\,gram$
So, the correct answer is option (A).
Additional information:
Another important parameter in the radioactive decay is half-life. The half life of the radioactive sample is defined as the time taken by the radioactive sample to decay to half of its initial quantity. It is denoted by the symbol ${t_{\dfrac{1}{2}}}$. The relation between half and decay constant is,
$\Rightarrow {t_{\dfrac{1}{2}}} = \dfrac{{\ln 2}}{\lambda }$

Note: In the radioactive decay equation, $N = {N_0}{e^{ - \lambda t}}$, $N$ represents total number of particles in the sample. If ${N_0}$ is the weight of the initial sample then N also represents the weight of the sample at time t.