Question

The activity of a radioactive sample is measured as $N_{0}$ counts per minute at t = 0 and $N_{0}/e$ counts per minute at t = 5 minutes. The time (in minutes) at which the activity reduces to half its value is

• A. $\log_{e}\frac{2}{5}$
• B. $\frac{5}{\log_{e}2}$
• C. $5\log_{e}2$
• D. $5\log_{e}2$

Answer 1

Answer: (D)

$5\log_{e}2$

Answer 2

: According to activity law

$R = R_{0}e^{- \lambda t}$……. (i)

Where

$R_{0}$= initial activity at t-0

R = activity at time t

$\lambda$ = decay constant

According to given problem

$R_{0} = N_{0}$counts per minute

$R = \frac{N_{0}}{e}$counts per minute

T =5 minutes

Substitution these values in equation (i), we get

$\frac{N_{0}}{e} = N_{0}e^{- 5\lambda}$

$e^{- 1} = e^{- 5\lambda}$

$5\lambda = 1or\lambda = \frac{1}{5}$ per minute

At t = $T_{1/2}$the activity R reduces to $\frac{R_{0}}{2}$

Where $T_{1/2}$= half life of a radioactive sample

From equation (i), we get

$\frac{R_{0}}{2} = R_{0}e^{- \lambda T_{1/2}}$

$e^{\lambda T_{1/2}} = 2$

Taking natural logarithms on both sides of above equation , we get

$\lambda T_{1/2} = \lg_{e}2$

Or $T_{1/2} = \frac{\log_{e}2}{\lambda} = \frac{\log_{e}2}{\left( \frac{1}{5} \right)}$

$= 5\log_{e}2$minutes