The mean life time of a radionuclide,if its a | vedclass

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(A) The activity $A$ of a radioactive sample is given by $A = |dN/dt| = \lambda N$.Given that the activity decreases by $4\%$ in $1 \ h$,we have the f

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$25$

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(A) The activity $A$ of a radioactive sample is given by $A = |dN/dt| = \lambda N$.Given that the activity decreases by $4\%$ in $1 \ h$,we have the fractional change in activity: $\frac{\Delta A}{A} = 0.04$ over $\Delta t = 1 \ h$.Since $A = A_0 e^{-\lambda t}$,for small $\lambda t$,we can approximate $A = A_0(1 - \lambda t)$,which gives $\frac{A_0 - A}{A_0} = \lambda t$.Substituting the given values: $0.04 = \lambda 	imes (1 \ h)$.Thus,the decay constant $\lambda = 0.04 \ h^{-1}$.The mean life time $T$ is the reciprocal of the decay constant: $T = \frac{1}{\lambda}$.$T = \frac{1}{0.04} \ h = 25 \ h$.

The mean life time of a radionuclide,if its activity decreases by $4\%$ for every $1 \ h$,would be .......... $h$ (product is non-radioactive,i.e.,stable).

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