A radioactive sample has an average life of 3 | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

$\ln \left(e^{-\lambda t}ight) =\ln \left(e^{-\frac{t}{R C}}ight)$

$-\lambda t =-\frac{t}{R C}$

$\lambda =\frac{1}{R C}$

$R =\frac{1}{\lambda C}$

$R =\frac{30 	imes 10^{-3}}{200 	imes 10^{-6}}$

$R =150 \Omega$

$\Rightarrow \lambda=\frac{1}{R C}$

$\Rightarrow R=\frac{1}{\lambda C}$

$\Rightarrow R=\frac{30 	imes 10^{-3}}{200 	imes 10^{-6}}$

$\Rightarrow R=150 \Omega$

Similar Questions

Certain radioactive substance reduces to $25\%$ of its value in $16\ days$. Its half-life is .......... $days$

At a given instant there are $25\%$ undecayed radioactive nuclei in a same. After $10 \,sec$ the number of undecayed nuclei reduces to $6.25\%$, the mean life of the nuclei is...........$ sec$

The half life of a radioactive element which has only $\frac{1}{{32}}$ of its original mass left after a lapse of $60\, days$ is ........$days$

The half life of a radioactive substance is $20$ minutes. In $........\,minutes$ time,the activity of substance drops to $\left(\frac{1}{16}\right)^{ th }$ of its initial value.

A radio isotope $X$ with a half-life $1.4 \times 10^{9}\; years$ decays of $Y$ which is stable. A sample of the rock from a cave was found to contain $X$ and $Y$ in the ratio $1: 7$. The age of the rock is ........ $ \times 10^9\; years$