Question

After 2 hours the amount of a certain radioactive substance reduces to \[\dfrac{1}{6}th\]of the original amount (the decay process follows first-order kinetics). The half-life of the radioactive substance is:
(a) 15min
(b) 30min
(c) 45min
(d) 60min

Answer 1

Hint: Radioactive decay depends on the initial concentration of the radioactive substance. Half-life of the reaction depends on the decay constant.

Complete answer:
Radioactive decay is the disintegration of an unstable atomic nucleus into smaller nuclei of atoms. It is a process where an unstable atom loses energy by radiation. The common types of radioactive decay are alpha decay, beta decay and gamma decay. All of these decays involve emitting one or more particles or photons. The rate of radioactive decay is directly proportional to the concentration (N) of the radioactive substance. It can be expressed as
\[ \dfrac{dN}{dt}=\lambda N\],here \[\lambda \]is the decay constant and \[\dfrac{dN}{dt}\]is the rate.
We know that radioactive decay is the first order reaction. So, its equation is as follows
\[k=\dfrac{2.303}{t}\log \dfrac{a}{a-x}\]
Here, k is the rate constant and in this case it is also decay constant \[\lambda \]. The initial concentration of the substance is a or \[{{N}_{0}}\] and after \[\dfrac{1}{6}th\] of the original amount, the concentration becomes N. So, we can write the equation in terms of radioactive decay constant. \[\lambda =\dfrac{2.303}{t}\log \dfrac{{{N}_{0}}}{N}\]
Here t is 2 hours=(2x60)min as given in the question and N= \[N=\dfrac{{{N}_{0}}}{6}\]
Let us substitute these values in above equation
\[\lambda =\dfrac{2.303}{2\times 60}\log \dfrac{{{N}_{0}}}{\dfrac{{{N}_{0}}}{6}}=\dfrac{2.303}{120\min }\log 6\]
\[\lambda =0.0149{{\min }^{-1}}\]
Now, half life of a first order reaction is given as the equation
\[{{t}_{\dfrac{1}{2}}}=\dfrac{0.693}{k}\]
For a radioactive decay, the half life of the decay is the time taken for half of the radioactive substance to decay and given as the equation
\[{{t}_{\dfrac{1}{2}}}=\dfrac{0.693}{\lambda }\]
Now substitute the value of decay constant we calculated.
\[{{t}_{\dfrac{1}{2}}}=\dfrac{0.693}{0.0149}\]
\[=46.4\min \simeq 45\min \]
Hence, the half life of the decay process or the radioactive decay of the radioactive substance is 45min.

The correct answer to the question is option (c).

Note: While substituting the value of N, we should know that the amount of the radioactive substance after 2 hours becomes \[\dfrac{1}{6}th\]of the initial amount. N here is not equal to \[\dfrac{1}{6}\].