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A radioactive sample has an average life of $30\, {ms}$ and is decaying. A capacitor of capacitance $200\, \mu\, {F}$ is first charged and later connected with resistor $^{\prime}{R}^{\prime}$. If the ratio of charge on capacitor to the activity of radioactive sample is fixed with respect to time then the value of $^{\prime}R^{\prime}$ should be $....\,\Omega$
$100$
$200$
$150$
$250$
Solution
$q=q_{0} e^{-\frac{t}{R C}}$
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
$\ln \left(e^{-\lambda t}\right) =\ln \left(e^{-\frac{t}{R C}}\right)$
$-\lambda t =-\frac{t}{R C}$
$\lambda =\frac{1}{R C}$
$R =\frac{1}{\lambda C}$
$R =\frac{30 \times 10^{-3}}{200 \times 10^{-6}}$
$R =150 \Omega$
$\Rightarrow \lambda=\frac{1}{R C}$
$\Rightarrow R=\frac{1}{\lambda C}$
$\Rightarrow R=\frac{30 \times 10^{-3}}{200 \times 10^{-6}}$
$\Rightarrow R=150 \Omega$