Question

Half-life of a substance is 10 years. In what time, it becomes \[\frac{1}{4}\,th\] part of the initial amount
A. 5 years
B. 10 years
C. 20 years
D. None of the above.

Answer 1

Hint: To find time for which radioactive substance becomes \[\frac{1}{4}\,th\] of initial amount use \[N = \frac{1}{4}\,{N_0}\] in equation \[N = {N_0}{e^{ - \lambda t}}\]Where, is decay constant and t is the amount of time in which substance becomes \[\frac{1}{4}\,th\] of initial amount.

Formula used:
\[N = {N_0}{e^{ - \lambda t}}\]
Here, N = Number of atoms after time t \[\,{N_0} = \]Initial number of atoms and \[\lambda = \] Decay Constant

Complete answer:
Given here is a radioactive substance with half life of 10 years, we have to find the time in which it decays to \[\frac{1}{4}\,th\] of initial amount.
Let the initial amount of the substance be and the time in which it becomes \[\frac{1}{4}\,th\] of initial amount be t.
As after time t substance becomes \[\frac{1}{4}\,th\] of initial amount,
Then, the amount of substance after t will be \[N = \frac{{{N_0}}}{4}\,\].
We know that, \[N = {N_0}{e^{ - \lambda t}}\,........(1)\]
Where, \[\lambda \]is Decay constant.
Substituting \[N = \frac{{{N_0}}}{4}\,\] in equation (1) we get,
\[\frac{{{N_0}}}{4} = {N_0}{e^{ - \lambda t}}\, \Rightarrow \frac{{{1}}}{4} = \,{e^{ - \lambda t}}\,.......(2)\]
Equation (2) can be simplified as,
\[{e^{\lambda t}} = 4\,.......(3)\]
 Decay constant is given by the equation,
\[\lambda = \frac{{\ln 2}}{T}\] Here, T is half life.
Substituting \[\lambda = \frac{{\ln 2}}{T}\] in equation (3) we get,
\[{e^{\frac{{\ln 2}}{T}t}} = 4\,.......(4)\]
Taking natural log both side of equation we get,
\(\frac{{\ln 2}}{T} \times t = \ln 4 \Rightarrow \frac{{\ln 2}}{T} \times t = 2\,\ln \,2\)
Further simplifying the above equation we get,
\(\frac{t}{T} = 2\,\,.......(4)\)
Substituting, T = 10 years in equation (4) we get,
\(\frac{t}{{10}} = 2\, \Rightarrow t = 20\,years\)
Hence, in 20 years the given substance will become \[\frac{1}{4}\]of the initial amount.
Therefore, option C is the correct option.



Note: This problem can be solved by using trial and error method for the time duration t given in options in equation \[N = {N_0}{e^{ - \lambda t}}\] and for the value of t for which equation is satisfied will be the correct option.