Question

After time equal to four half-lives, the amount of radioactive material remaining undecayed is-
(A) 6.25 %
(B) 12.50 %
(C) 25.00 %
(D) 50.00 %

Answer 1

Hint:A heavy nucleus is unstable therefore it becomes stable by emitting various particles viz, alpha, beta particles, electrons, etc. The radioactive decay is exponential i.e. it takes infinite time to fully decay. Half-live is the time after which the final amount is half of the initial amount of radioactive substance.

Formula used:
\[
\Rightarrow N = {N_0}\exp ( - \lambda t) \\ and \\
\Rightarrow N = {N_0}{\left( {\dfrac{1}{2}} \right)^n} \\
\]
Where,
${N_0}$ is the amount of radioactive substance at time t=0
n is the number of half-lives
$\lambda $ is the decay constant

Complete step by step answer:
Radioactivity was first discovered by Henri Becquerel in 1896 when he observed that some phosphorescent materials glow in dark. Radioactivity is defined as the method by which heavy unstable large nuclei become smaller stable nuclei by emitting certain particles like alpha, beta, electrons, etc. Mathematically it is described as,
\[\Rightarrow N = {N_0}\exp ( - \lambda t)\]
The above expression states that at time t=0, the number of particles is ${N_0}$ and it decreases exponentially over time. A half life of the material is defined as the time taken to disintegrate to half of its initial amount (number of particles). After one half life,
\[
\Rightarrow N = \dfrac{{{N_0}}}{2} \\
\Rightarrow {t_{\dfrac{1}{2}}} = \dfrac{{\ln 2}}{\lambda } \\
\]
So, after n half-lives, the number of particles remaining (undecayed) will be,
\[\Rightarrow N = {N_0}{\left( {\dfrac{1}{2}} \right)^n}\]
So, after 4 half-lives the number of particles remaining is:
\[\Rightarrow N = {N_0}{\left( {\dfrac{1}{2}} \right)^4} = \dfrac{{{N_0}}}{{16}}\]
In percentage,
\[
\Rightarrow \dfrac{{final}}{{initial}} \times 100 \\
\Rightarrow \dfrac{{\dfrac{{{N_0}}}{{16}}}}{{{N_0}}} \times 100 = 6.25\% \\
\]
Therefore, the amount of undecayed radioactive substance is 6.25% of the original amount.
The correct answer is option A.

Note: To find the decayed amount of radioactive substance, subtract the remaining undecayed amount from the original amount at time t=0 i.e,
\[
\Rightarrow {N_{decay}} = {N_0} - N = {N_0} - {N_0}{\left( {\dfrac{1}{2}} \right)^n} \\
\Rightarrow {N_{decay}} = {N_0}\left( {1 - \dfrac{1}{{{2^n}}}} \right) \\
\]