A radioactive sample has an average life of 3 | vedclass

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Question

$q=q_{0} e^{-\frac{t}{R C}}$

where, $R C$ is the time constant of $R C$ - circuit.

At time $t$, activity $A$ is given by

$A=A_{0} e ^{-\lambd

Vedclass · Accepted Answer

$150$

Vedclass · Answer

$q=q_{0} e^{-\frac{t}{R C}}$

where, $R C$ is the time constant of $R C$ - circuit.

At time $t$, activity $A$ is given by

$A=A_{0} e ^{-\lambda t} $ where, $A_{0}=$ initial activity and $\quad \lambda=$ decay constant. On dividing Eqs. (i) and (ii), we get

$\frac{q}{A} =\frac{q_{0} e^{-\frac{t}{R C}}}{A_{0} e^{-\lambda t}}$

$1=\frac{e^{-\frac{t}{R C}}}{e^{-\lambda t}}$

$e^{-\lambda t} =e^{-\frac{t}{R C}}$

$\Rightarrow$ Taking log on both sides of above equation, we get

$\ln \left(e^{-\lambda t}ight) =\ln \left(e^{-\frac{t}{R C}}ight)$

$-\lambda t =-\frac{t}{R C}$

$\lambda =\frac{1}{R C}$

$R =\frac{1}{\lambda C}$

$R =\frac{30 	imes 10^{-3}}{200 	imes 10^{-6}}$

$R =150 \Omega$

$\Rightarrow \lambda=\frac{1}{R C}$

$\Rightarrow R=\frac{1}{\lambda C}$

$\Rightarrow R=\frac{30 	imes 10^{-3}}{200 	imes 10^{-6}}$

$\Rightarrow R=150 \Omega$

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