Question

The half life of Carbon-14 ( $ {C^{14}} $ ) is $ 5715 $ year. What is its decay constant?

Answer 1

Hint
The relation between half-life and decay constant is $ {t_{{1}/{2}}} = \dfrac{{0.693}}{\lambda } $ (where, $ \lambda $ is the decay constant). The decay constant is the reciprocal of time during which the number of atoms of a radioactive substance decreases to 1/e (or, 36.8%) of the number present initially.

Complete step by step answer
 For a radioactive sample, neither can it be predicted which nucleus will disintegrate first, nor can the sequence of occurrence be ascertained beforehand. Only we can say that the time rate of disintegration will be directly proportional to the number of radioactive particles present in the sample at that time. Let, at the time $ t $ , the number of radioactive particles present in the sample be $ N $ .
So, $ \dfrac{{dN}}{{dt}} \propto N $ or, $ \dfrac{{dN}}{{dt}} = - \lambda N $ (where, $ \lambda $ is the decay constant).
So, the decay constant is the reciprocal of time during which the number of atoms of a radioactive substance decreases to 1/e (or, 36.8%) of the number present initially.
Now, the time period after which the number of radioactive atoms present in a radioactive sample becomes half of its initial number due to disintegration is called half-life of that radioactive element.
The relation between half life and decay constant is $ {t_{{1}/{2}}} = \dfrac{{0.693}}{\lambda } $
Given that the half life of Carbon-14 is $ 5715 $ year.
So, $ \lambda = \dfrac{{0.693}}{{5715}} = 0.0001212598 $ .
So, the decay constant is $ 0.0001212598 $ .

Note
Let, at the beginning of the count for disintegration, i.e., at $ t = 0 $ number of radioactive atoms present in a radioactive sample = $ {N_0} $ . After a time $ t $ this number becomes $ N $ .
According to the exponential law, $ N = {N_0}{e^{ - \lambda t}} $ [where $ \lambda $ is the decay constant].
Now, if the half-life of that element is $ T $ and then after time $ T $ the number of atoms present in the sample $ N $ becomes $ \dfrac{{{N_0}}}{2} $ ,
 $ \dfrac{{{N_0}}}{2} = {N_0}{e^{ - \lambda T}} $ or, $ {e^{\lambda T}} = 2 $ or, $ \lambda T = {\log _e}2 $
or, $ T = \dfrac{{{{\log }_e}2}}{\lambda } = \dfrac{{2.303{{\log }_{10}}2}}{\lambda } = \dfrac{{0.693}}{\lambda } $