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
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
Updated On: Aug 15, 2022
- $\frac{A_0}{4}$
- $\frac{A_0}{10}$
- $\frac{A_0}{15}$
- $\frac{A_0}{25}$
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The Correct Option is D
Solution and Explanation
$\because$ 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)]
$\therefore A =\frac{ A _{0}}{5 \times 5}=\frac{ A _{0}}{25}$
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