Question

At a given instant, there are \[25\% \] undecayed radioactive nuclei in a sample. After \[10\,s\] the number of undecayed nuclei reduces to \[12.5\% \] . Calculate
(a) mean life of the nuclei and
(b) the time in which the number of decayed nuclei will further reduce to \[6.25\% \] of the reduced number.

Answer 1

Hint: We are asked to find two things in the question, the time at which a particular value occurs and mean life of the nuclei. We start by writing down the values and data given in the question. The mean life of the nuclei can be found by using a direct formula. Then moving onto the time in which the number of decayed nuclei will further reduce to \[6.25\% \] , we find the number of half-lives it takes to decay to this value and multiply it with the given time.

Formulas used:
The formula to find the mean life is given by,
\[\tau = \dfrac{T}{{{{\log }_e}2}}\]
The formula to find the number of half lives is given by,
\[\dfrac{N}{{100}} = {\left( {\dfrac{1}{2}} \right)^n}\]
Where \[n\] is the number of half lives, \[N\] is the particular amount of decay that has happened in fraction and \[T\] is the time taken for one half life.

Complete step by step answer:
Let us start by noting the data given in the above question, the time taken by half life to happen is, \[T = 10\,s\]. The particular amount of decay that has happened in fraction is, \[N = 6.25\].

(a) Now that we have the values given, let us move onto the first part of the question, which is to find the mean life of the nuclei. This can be done by directly applying the formula,
\[\tau = \dfrac{T}{{{{\log }_e}2}}\]
Substituting the values, we get
\[\tau = \dfrac{T}{{{{\log }_e}2}} \\
\Rightarrow \tau = \dfrac{{10}}{{0.693}} \\
\therefore \tau = 14.43\,s\]
Now that we have the value of mean life, let us move onto the second part of the question and see how many half-lives are crossed in the given particular decay amount

(b) We use the formula \[\dfrac{N}{{100}} = {\left( {\dfrac{1}{2}} \right)^n}\] to find the number of half-lives crossed
\[\dfrac{{6.25}}{{100}} = {\left( {\dfrac{1}{2}} \right)^n} \\
\Rightarrow \dfrac{1}{{16}} = {\left( {\dfrac{1}{2}} \right)^n}\]
To find the value of \[n\] we take the reciprocal and see two powered what is sixteen
\[{2^n} = 16 \\
\Rightarrow {2^4} = 16\]
This means that the number of half lives crossed is, \[n = 4\]. Now we find the time taken by taking the product of the number of half lives taken and the time taken for one half life,
\[t = 4 \times T \\
\Rightarrow t= 4 \times 10 \\
\therefore t= 40s\]

Therefore, the mean life of the nuclei will be \[14.43s\] and the time taken for \[6.25\% \] decay is, \[40s\].

Note: The value of \[n\] can also be found taking the logarithm function on both sides, that is, \[\log \left( {16} \right) = n\log \left( 2 \right)\]
We use the property \[\log {\left( a \right)^n} = n\log \left( a \right)\]
The value of \[n\] can be found as \[n = 4\] when we divide the two values.