Question

A radioactive material has a half-life of 8 years. The activity of the material will decrease to about \[\dfrac{1}{8}\] of its original value in
A. 256 years
B. 128 years
C. 64 years
D. 24 years

Answer 1

Hint:The half-life of a radioactive substance is the amount of time it takes for its concentration to decrease by half. Keep in mind that if the half-life is short, the radioactive chemical has a high activity, which denotes that it is highly radioactive and should be handled carefully.

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

Complete step by step solution:
To find the time taken for \[\dfrac{1}{8}\] of the substance to decay, we must rearrange the above two formulae as,
\[N = {N_0}{e^{ - \dfrac{{\ln 2}}{{{T_{\dfrac{1}{2}}}}}t}}\]
\[\Rightarrow N = {N_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{{T_{\dfrac{1}{2}}}}}}}\]
Here, half-life is given as \[{T_{\dfrac{1}{2}}}\] = 8 years and \[N = \dfrac{1}{8}{N_0}\]
So, substituting it in the above equation, we get,
\[\dfrac{{{N_0}}}{8} = {N_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{8}}} \\ \]
\[\Rightarrow \dfrac{1}{8} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{8}}} \\ \]
\[\Rightarrow {\left( {\dfrac{1}{2}} \right)^3} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{8}}}\]
Now, the bases are the same in the above equation, then we can equate the powers raised against them.
\[3 = \dfrac{t}{8}\]
\[\therefore t = 24\,years\]

Hence, the correct answer is option D.

Note: The alternate method to clear this question is to consider the number of half-lives the radioactive substance has to undergo to reduce to \[\dfrac{1}{8}\] of its original amount. Consider the following,
\[{N_0}\mathop \to \limits^{1{T_{\dfrac{1}{2}}}} \mathop {\dfrac{{{N_0}}}{2}}\limits_{\left( {0.5{N_0}} \right)} \mathop \to \limits^{2{T_{\dfrac{1}{2}}}} \mathop {\dfrac{{{N_0}}}{4}}\limits_{\left( {0.25{N_0}} \right)} \mathop \to \limits^{3{T_{\dfrac{1}{2}}}} \mathop {\dfrac{{{N_0}}}{8}}\limits_{\left( {0.125{N_0}} \right)} \\ \]
It takes 3 half-lives for the original amount to reduce to 1/8 of its initial amount. Since one half-life is 8 years, then 3 half-lives will be 24 years.