A radioactive sample decays with an average-life of $20 m s$ . A capacitor of capacitance $100 μ F$ is charged to some potential and then the plates are connected through a resistance $R$ . What should be the value of $R$ (in $\times 10^{2} Ω$ ) so that the ratio of the charge on the capacitor to the activity of the radioactive sample remains constant in time?

Question

Verified by Toppr

Answer 1

The activity of the sample at time $t$ is given by $A = A_{0} e^{- λ t}$ where $λ$ is the decay constant and $A_{0}$ is the activity at time $t = 0$ when the capacitor plates are connected. The charge on the capacitor at time $t$ is given by
$Q = Q_{0} e^{- t / C R}$
where $Q_{0}$ is the charge at $t = 0$ and $C = 100 μ F$ Thus
$\frac{Q}{A} = \frac{Q _{0}}{A _{0}} \frac{e ^{- t / C R}}{e ^{- λ t}}$
It is independent of it if $λ = \frac{1}{C R}$
$R = \frac{1}{λ C} = \frac{t _{a v}}{C}$
$= \frac{2 0 \times 1 0 ^{- 3} s}{1 0 0 \times 1 0 ^{- 6} F} = 200 Ω$

Answer 2

The activity of the sample at time

t

is given by

A = A_{0} e^{- λ t}

where

λ

is the decay constant and

A_{0}

is the activity at time

t = 0

when the capacitor plates are connected. The charge on the capacitor at time

t

is given by

Q = Q_{0} e^{- t / C R}

where

Q_{0}

is the charge at

t = 0

and

C = 100 μ F

Thus

\frac{Q}{A} = \frac{Q _{0}}{A _{0}} \frac{e ^{- t / C R}}{e ^{- λ t}}

It is independent of it if

λ = \frac{1}{C R}

R = \frac{1}{λ C} = \frac{t _{a v}}{C}

= \frac{2 0 \times 1 0 ^{- 3} s}{1 0 0 \times 1 0 ^{- 6} F} = 200 Ω