Question

A block having $12\;g$ of an element is placed in a room. This element is a radioactive element with a half-life of $15\;$ years. After how many years will there be just $1.5\;g$ of the element in the box?
(A) $40\;$ years
(B) $45\;$ years
(C) $20\;$ years
(D) $15\;$ years

Answer 1

Hint: The decay of a radioactive element is exponential, the half-life indicates the time taken by the element to reduce to half of its previous amount. We can first determine the decay constant by going through the definition of half-life and then put it in the formula for the exponential decay.
Formula used:
${e^{ - \lambda t}} = \dfrac{N}{{{N_0}}}$

Complete step by step answer:
It is given in the question that,
The initial mass of the element, ${N_0} = 12\;g$
The half-life of the element, ${t_{\dfrac{1}{2}}} = 15\;years$
We know that the half-life of an element is given by,
${t_{\dfrac{1}{2}}} = \dfrac{{\ln 2}}{\lambda }$
where $\lambda $ is the decay constant.
To find $\lambda $ we rearrange the equation and keep the value of $\ln 2$ as $0.693\;$
Therefore,
$\lambda = \dfrac{{0.693}}{{15}}$
$ \Rightarrow \lambda = 0.0462$
The exponential decay of an element is given by,
${e^{ - \lambda t}} = \dfrac{N}{{{N_0}}}$
where $t$ is the time elapsed,
$N$ is the amount of element that is left,
${N_0}$ is the initial amount of element,
$\lambda $ is the decay constant.
It is given in the question that,
$N = 1.5g$
$\Rightarrow$ ${N_0} = 12g$
$\Rightarrow$ $\lambda = 0.0462$
Putting these values in the formula,
${e^{ - \lambda t}} = \dfrac{N}{{{N_0}}}$
$ \Rightarrow {e^{ - 0.0462t}} = \dfrac{{1.5}}{{12}}$
Taking logarithm on both sides,
$\Rightarrow$ $\ln ({e^{ - 0.0462t}}) = \ln \left( {\dfrac{{1.5}}{{12}}} \right)$
Since,
$\ln {x^n} = n\ln x$
The equation can be written as,
$ - 0.0462t = \ln \left( {0.125} \right)$
From the log table,
$\ln (0.125) = - 2.0794$
Using this value,
$t = \dfrac{{ - 2.0794}}{{ - 0.0462}}$
$\Rightarrow$ $t = 45$

Therefore the time taken is equal to $45\;$ years, this makes option (B) the correct answer.

Note: An alternate solution can be that the half-life is a widely used term to define the stability of a radioactive substance. To avoid lengthy calculations, we can compare the number of half-lives of the element passed with the amount left. Like if we repeatedly divide $12\;$ by $2$ we find that on the third time the amount becomes $1.5\;$ , this means that it takes three half-lives for the element to decay by this amount, and $3 \times {t_{\dfrac{1}{2}}}$ is $45\;$ years. This method can be used in competitive exams with multiple-choice questions that do not require a full solution.