Question

A sample of a radioactive element has a mass of 10g at an instant t = 0. The approximate mass of this element in the sample after two mean lives is
A.) 1.35g
B.) 2.50g
C.) 3.70g
D.) 6.30g

Answer 1

Hint: A radioactive material exhibits exponential decay in its quantity. The rate of decay depends on the amount of radioactive mass available in the sample at an instant. Mean life of the element is equal to the reciprocal of the decay constant.

Complete step-by-step answer:
The law of radioactive decay is given as follows:

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

where N is the number of nuclides after time t and ${N_0}$ is the number of nuclides at t = 0. This relation describes the exponential decay of a sample containing radioactive material. It depends on the number of nuclides present in the sample at a particular instant of time. The number of nuclides can also be replaced by other quantities which change with decay of the sample (like mass).

$\lambda$ is the decay constant of the element and signifies the probability of decay of the sample per unit time.

Mean life of a radioactive sample is the average life-time of the nuclei in the sample. It is equal to the reciprocal of the decay constant of the element which is equal to the half-life of the element divided by natural logarithm of 2.

$\tau = \dfrac{1}{\lambda } = \dfrac{{{T_{\dfrac{1}{2}}}}}{{\ln 2}}$

We are given that at t = 0, the mass of the radioactive sample is 10g.
$\therefore {m_0} = 10g$

Now we can write the decay equation in terms of mass as
$m = {m_0}{e^{ - \lambda t}}$

We are given that t = 2 mean lives = $2\tau = \dfrac{2}{\lambda }$

$\begin{gathered} \therefore m = 10{e^{ - \lambda \times \dfrac{2}{\lambda }}} \\\ \Rightarrow m = 10{e^{ - 2}} \\\ \Rightarrow m = \dfrac{{10}}{{{e^2}}} \\\ \Rightarrow m = 1.35g \\\ \end{gathered}$

So, the correct answer is option A.

Note: The decay rate of the radioactive material is a statistical quantity which means that it depends on the concentration of radioactive nuclides in the radioactive sample and not on the individual nuclei.