A radioactive sample is undergoing \[\alpha \]decay. At any time \[{t_1}\], its activity is A and at another time \[{t_2}\], the activity is \[\dfrac{A}{5}\] . What is the average lifetime for the sample?
A. \[\dfrac{{\left( {{t_2} - {t_1}} \right)}}{{\ln 5}}\]
B. \[\dfrac{{\ln \left( {{t_2} + {t_1}} \right)}}{2}\]
C. \[\dfrac{{\left( {{t_1} - {t_2}} \right)}}{{\ln 5}}\]
D. \[\dfrac{{\ln 5}}{{\left( {{t_2} - {t_1}} \right)}}\]

Hint:Radioactive decay is defined as the emission of energy in the form of ionizing radiation. It involves the spontaneous transformation of one element into another. This can happen only by changing the number of protons in the nucleus.

Formula Used:
To find the activity of the radioactive sample the formula is,
\[A = {A_0}{e^{ - \lambda t}}\]
Where, \[{A_0}\] is initial activity, \[\lambda \] is decay constant and \[t\] is the half-life of a decaying substance.

Complete step by step solution:
Consider a radioactive sample which undergoes \[\alpha \] decay. At any time \[{t_1}\], its activity is A and at another time \[{t_2}\], the activity is \[\dfrac{A}{5}\] . We need to find the average lifetime for the sample. We have studied that the activity of the radio sample is,
\[A = {A_0}{e^{ - \lambda t}}\]
That is the activity decreases exponentially with time.
Here, the activity at time \[{t_1}\]is,
\[{A_1} = {A_0}{e^{ - \lambda {t_1}}}\]……….. (1)

Similarly, the activity at time \[{t_2}\]is,
\[{A_2} = {A_0}{e^{ - \lambda {t_2}}}\]………….. (2)
Here they have given, \[{A_1} = A\]and \[{A_2} = \dfrac{A}{5}\]
Then, the equation (1) and (2), we get,
\[A = {A_0}{e^{ - \lambda {t_1}}}\]……… (3)
\[\Rightarrow \dfrac{A}{5} = {A_0}{e^{ - \lambda {t_2}}}\]……….. (4)

Now, divide the equation (3) by (4), we get,
\[\dfrac{A}{{\dfrac{A}{5}}} = \dfrac{{{A_0}{e^{ - \lambda {t_1}}}}}{{{A_0}{e^{ - \lambda {t_2}}}}}\]
\[\Rightarrow 5 = \dfrac{{{e^{ - \lambda {t_1}}}}}{{{e^{ - \lambda {t_2}}}}}\]
\[\Rightarrow 5 = {e^{\lambda \left( {{t_2} - {t_1}} \right)}}\]
To eliminate the exponential term, we will take the natural logarithm on both sides, that is,
\[\lambda \left( {{t_2} - {t_1}} \right) = \ln 5\]
\[\Rightarrow \dfrac{1}{\lambda } = \dfrac{{\left( {{t_2} - {t_1}} \right)}}{{\ln 5}}\]

The mean or average lifetime is,
\[\tau = \dfrac{1}{\lambda }\]
\[\therefore \tau = \dfrac{{\left( {{t_2} - {t_1}} \right)}}{{\ln 5}}\]
Therefore, the average lifetime for the sample is \[\dfrac{{\left( {{t_2} - {t_1}} \right)}}{{\ln 5}}\].

Hence, Option A is the correct answer

Note:Remember that whenever there is an exponential term, in order to solve this, we need to take the natural logarithm on both sides, then we can resolve it easily.