Question

The half-life of a radioactive sample is 2x years. What fraction of this sample remains undecayed after x years?
(A) $\dfrac{1}{2}$
(B) $\dfrac{1}{{\sqrt 2 }}$
(C) $\dfrac{1}{{\sqrt 3 }}$
(D) $2$

Answer 1

Hint : Half life of a radioactive compound or any isotope is the time taken by that isotope to reduce to half its original value. To find how much of the compound that is un-decayed is left we will apply the concept that rate of decay is directly proportional to concentration of nuclei at that instant.
$\lambda = \dfrac{{2.303}}{t}\log \left( {\dfrac{{{N_0}}}{N}} \right) - - - - - - - (1)$
Where, $\lambda$ = half life , t= years given and $\dfrac{{{N_0}}}{N}$ is the fraction of decayed compound .

Complete Step By Step Answer:
Radioactive decay is the spontaneous breakdown of an atomic nucleus which results in the release of a lot of energy. It follows first order kinetics.
Given:
Half-life= $\lambda$ = 2x years and t=x years.
We also know that for first order reaction half-life is given by :
$\lambda = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}$ where ${t_{\dfrac{1}{2}}}$ is time for half-life.
Now let us substitute these values in equation 1,
$\lambda = \dfrac{{2.303}}{t}\log \left( {\dfrac{{{N_0}}}{N}} \right)$
Or, $\dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}} = \dfrac{{2.303}}{t}\log \left( {\dfrac{{{N_0}}}{N}} \right)$
Or, $\dfrac{{0.693}}{{2x}} = \dfrac{{2.303}}{x}\log \left( {\dfrac{{{N_0}}}{N}} \right)$
Or, $\dfrac{{0.693}}{{2 \times 2.303}} = \log \left( {\dfrac{{{N_0}}}{N}} \right)$
Or, $\dfrac{{0.693}}{{4.606}} = \log \left( {\dfrac{{{N_0}}}{N}} \right)$
Or, $0.150 = \log \left( {\dfrac{{{N_0}}}{N}} \right)$
Or, $\dfrac{1}{2}\log 2 = \log \left( {\dfrac{{{N_0}}}{N}} \right)$ (0.150 can also be written as log2 divided by 2),
Or, $\log \sqrt 2 = \log \left( {\dfrac{{{N_0}}}{N}} \right)$ ( because $alogb = {\text{ }}log{b^a}$ )
Or, $\left( {\dfrac{{{N_0}}}{N}} \right) = \sqrt 2$
Hence the fraction of undecayed compound will be : $\left( {\dfrac{N}{{{N_0}}}} \right) = \dfrac{1}{{\sqrt 2 }}$
Hence we see that the fraction of compounds left un-decayed is $\dfrac{1}{{\sqrt 2 }}$ . Hence the correct answer to this question is option B.

Note :
Remember that fraction $\left( {\dfrac{N}{{{N_0}}}} \right)$ is the fraction for undecayed compound and not $\left( {\dfrac{{{N_0}}}{N}} \right)$ . If you take the latter as the fraction answer will be wrong. Also note that radioactivity follows 1st order kinetics hence its formula for half-life is $\lambda = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}$ . Do not apply the formula for half-life of zero-order reaction.