13.Nuclei
hard

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 $....\,\Omega$

A

$100$

B

$200$

C

$150$

D

$250$

(JEE MAIN-2021)

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

$\ln \left(e^{-\lambda t}\right) =\ln \left(e^{-\frac{t}{R C}}\right)$

$-\lambda t =-\frac{t}{R C}$

$\lambda =\frac{1}{R C}$

$R =\frac{1}{\lambda C}$

$R =\frac{30 \times 10^{-3}}{200 \times 10^{-6}}$

$R =150 \Omega$

$\Rightarrow \lambda=\frac{1}{R C}$

$\Rightarrow R=\frac{1}{\lambda C}$

$\Rightarrow R=\frac{30 \times 10^{-3}}{200 \times 10^{-6}}$

$\Rightarrow R=150 \Omega$

Standard 12
Physics

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