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.