Question

A radioactive sample can decay by two different processes. The half-life for the first process is \[T_1 \] and that for the second process is \[T_2 \] . What will be the effective half life of the radioactive sample?

Answer 1

Hint: Use the relation between decay constant
Also half-life to substitute for decay constant in the radioactive decay equation.

Formula used:
\[T_{\dfrac{1}{2}} = \dfrac{{\ln 2}}{\lambda }\]
Here, \[T_{\dfrac{1}{2}} \] is the half-life and \[\lambda \] is the decay constant.

Complete step by step answer:
A rate of radioactive decay of a sample is directly proportional to the actual number of particles remaining in the substance at that instant of time t.
\[\dfrac{{dN}}{{dt}} = - \lambda N\]

Here, \[\lambda \] is the proportionality constant and it is known as decay constant. The negative sign represents the substance undergoes decay over a course of time t.

The term half-life represents the time required for a half of the nuclei in the sample to undergo the decay.

The parameter half-life is related to the decay constant \[\lambda \]by the equation,\[T_{\dfrac{1}{2}} =\dfrac{{\ln 2}}{\lambda }\]

\[ \Rightarrow \lambda = \dfrac{{\ln 2}}{{T_{\dfrac{1}{2}} }}\]

Since the radioactive sample decay by two processes, the effective rate of decay is,
\[\dfrac{{dN}}{{dt}} = - \lambda N\] ...... (1)

Here, \[\lambda = \lambda _1 + \lambda _2 \].
Therefore,
\[\dfrac{{dN}}{{dt}} = - \left( {\lambda _1 + \lambda _2 } \right)N\] ...... (2)

Compare equation (1) and (2), we can write,
\[ - \left( {\lambda _1 + \lambda _2 } \right)N = - \lambda N\]

\[ \Rightarrow \left( {\lambda _1 + \lambda _2 } \right)N = \lambda N\] ...... (3)

The decay constant for the first process is,
\[\lambda _1 = \dfrac{{\ln 2}}{{T_1 }}\] ...... (4)
And, the decay constant for the second process is,
\[\lambda _2 = \dfrac{{\ln 2}}{{T_2 }}\] ...... (5)

The effective decay constant for the two processes is,
\[\lambda = \dfrac{{\ln 2}}{T}\] ...... (6)

Substitute equations (4), (5), and (6) in equation (3).
\[\left( {\dfrac{{\ln 2}}{{T_1 }} + \dfrac{{\ln 2}}{{T_2 }}} \right)N = \dfrac{{\ln 2}}{T}N\]
\[ \Rightarrow \dfrac{1}{{T_1 }} + \dfrac{1}{{T_2 }} = \dfrac{1}{T}\]

So, the correct answer is “Option B”.

Note:
sometimes students do not consider the negative sign in radioactive decay equation \[
\dfrac{{dN}}{{dt}} = - \lambda N\].
It will affect the answer in other problems on radioactive decay.