Question

The activity of a radioactive element decreases in 10 years to 1/5 of initial activity $A_0$ . After further next 10 years its activity will be

• A. $\frac{A_0}{4}$
• B. $\frac{A_0}{10}$
• C. $\frac{A_0}{15}$
• D. $\frac{A_0}{25}$

Answer 1

Answer: (D)

$\frac{A_0}{25}$

Answer 2

$\because$ Activity of a radioactive sample, $A = A _{0}\left(\frac{1}{2}\right)^{\frac{ t }{ t _{1} / 2}}$ where, $A _{0}=$ initial activity. In first case, $\frac{ A _{0}}{5}= A _{0}\left(\frac{1}{2}\right)^{\frac{10}{t_{1} / 2}}$ $\frac{1}{5}=\left(\frac{1}{2}\right)^{\frac{10}{t_{1} / 2}}$...(i) In second case, $A = A _{0}\left(\frac{1}{2}\right)^{\frac{20}{ t _{1} / 2}}$...(ii) From Eqs. (i) and (ii), we get $\frac{1 / 5}{A}=\frac{\left(\frac{1}{2}\right)^{\frac{10}{t_{1} / 2}}}{A_{0}\left(\frac{1}{2}\right)^{\frac{20}{t_{1} / 2}}}$ $\Rightarrow \frac{1}{5 A}=\frac{1}{A_{0}\left(\frac{1}{2}\right)^{\frac{10}{t_{1} / 2}}}$ $\Rightarrow \frac{1}{5 A }=\frac{1}{\frac{ A _{0}}{5}}$ [from E (i)] $\therefore A =\frac{ A _{0}}{5 \times 5}=\frac{ A _{0}}{25}$