This question was previously asked in

Bihar CET B.Ed. Official Paper (Held on 16 Jan 2016)

Option 2 : 12.5%

__Concept:__

The half-life of a radioactive element (T1/2): time interval in which the mass of a radioactive substance or the number of atoms is reduced to half of its initial value.

- The expression for the half-life is

\(\Rightarrow {T_{\frac{1}{2}}} = \frac{{0.693}}{\lambda }\)

Where λ = is the decay rate constant

- If N0 is the initial concentration then after n half-lives the amount of substance(N) left is given by

\(\Rightarrow \frac{N}{N_{0}} = (\frac{1}{2})^{n}\)

__Calculation:__

Given:

\(T_{\frac{1}{2}} = 60 min = 1 Hour\) , t = 3 hours

- The number of half-life for t amount of time is given by

\(\Rightarrow n = \frac{T _{\frac{1}{2}}}{t } =\frac{3}{1} = 3 \)

The amount of substance remains after 3 half-lives are given by

\(\Rightarrow\frac{N}{N_{0}} = (\frac{1}{2})^{n}\)

\(\Rightarrow \frac{N}{N_{0}} = (\frac{1}{2})^{3}\)

\(\Rightarrow N =N_{0}\times \frac{1}{8}\)

- The amount left after 3 hours

\(\Rightarrow N = N_{0} -\frac{N_{0}}{8} = \frac{7N_{0}}{8}\)

Let N0 be the initial amount then the percentage change is given by

\(\Rightarrow \%\, decay=\frac{N}{N_o}=\frac{7N_o}{8}\times 100 = 87.5\%\)

So, the quantity left after 3 hours:

100 - 87.5% = **12.5%**