Question

A radioactive material has a half-life of 10 days. What fraction of the material would remain after 30 days?
A. 0.5
B. 0.25
C. 0.125
D. 0.33

Answer 1

Hint:The radioactive activity of a radioactive substance can be expressed by defining the half-life of the radioactive substance, because the lower the half-life of the radioactive substance the higher the activity of it. The half-life can be explained as the time desired for the radioactive element to decay into one half of its initial amount.

Formula used :
The half-life is given by,
\[{T_{\dfrac{1}{2}}} = \dfrac{{\ln 2}}{\lambda } = \dfrac{{0.6931}}{\lambda }\]
Where, \[\lambda \] - decay or disintegration constant
The law of radioactive decay,
\[N = {N_0}{e^{ - \lambda t}}\]
Where, N = number of atoms at time t
\[{N_0}\] = number of atoms at \[t = 0\]

Complete step by step solution:
Given, \[{T_{{\textstyle{1 \over 2}}}}\] = 10 days
t = 30 days
To find, N = ?
We should rearrange the above two formulae as such,
\[N = {N_0}{e^{ - \dfrac{{\ln 2}}{{{T_{\dfrac{1}{2}}}}}t}} \\ \]
\[\Rightarrow N = {N_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{{T_{\dfrac{1}{2}}}}}}}\]
(Since exponent and log are inverse functions, they get cancelled.)
Substituting the known values in the above equation,
\[N = {N_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{{30}}{{10}}}} \\ \]
\[\Rightarrow \dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^3} \\ \]
\[\Rightarrow \dfrac{N}{{{N_0}}} = \dfrac{1}{8} \\ \]
\[\therefore \dfrac{N}{{{N_0}}} = 0.125\]

Hence, the correct answer is option C.

Note :This can also be solved using another simple method, by calculating the half-lives of the radioactive substance until the given time period. Given, \[{T_{{\textstyle{1 \over 2}}}}\]= 10 days and the given time period, t = 30 days, which means three half-lives happened.
\[{N_0}\mathop \to \limits^{1{T_{\dfrac{1}{2}}}} \mathop {\dfrac{{{N_0}}}{2}}\limits_{\left( {0.5{N_0}} \right)} \mathop \to \limits^{2{T_{\dfrac{1}{2}}}} \mathop {\dfrac{{{N_0}}}{4}}\limits_{\left( {0.25{N_0}} \right)} \mathop \to \limits^{3{T_{\dfrac{1}{2}}}} \mathop {\dfrac{{{N_0}}}{8}}\limits_{\left( {0.125{N_0}} \right)} \]
Hence, 0.125 % of \[{N_0}\] remains after 30 days.