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NUCLEAR PHYSICS

148067 The activity of a radioactive element decreases to one-third of the original activity $A_{0}$ in a period of $9 yr$ . After a further lapes of $9 yr$ , its activity will be

A_{0}

\frac{2}{3} A_{0}

\frac{A_{0}}{9}

\frac{A_{0}}{6}

\frac{A_{0}}{18}

Explanation:

C Activity of radioactive elements given by -
$A = A_{o} e^{- λ t}$
According to question -
$A = \frac{A_{o}}{3}, t = 9 years$
$∴ \frac{A_{o}}{3} = A_{o} e^{- 9 λ}$
$\frac{1}{3} = e^{- 9 λ}$
When, $t = 18$ years
Then,
$A^{'} = A_{o} e^{- 18 λ}$
$A^{'} = A_{o} {(e^{- 9 λ})}^{2}$
From equation (i) and (ii) we get -
$A^{'} = A_{o} {(\frac{1}{3})}^{2}$
$A^{'} = \frac{A_{o}}{9}$

Kerala CEE - 2009

NEET DIGITAL LIBRARY