The activity of a radioactive sample is measured as N 0 counts per minute at time t = 0 and e N 0 counts per minute at time t = 3 minute. The time (in minute) in which the activity reduces to ha

Question

The activity of a radioactive sample is measured as N 0 ​ counts per minute at time t = 0 and e N 0 ​ ​ counts per minute at time t = 3 minute. The time (in minute) in which the activity reduces to half the value, is

TestHub · Accepted Answer

The activity of a radioactive sample at time $t$ is given by $A = A_{0} e^{- λ t}$ .

Given that at $t = 0$ , $A = N_{0}$ .

At $t = 3$ minutes, $A = \frac{N _{0}}{e}$ .

Substituting these values into the decay equation:

$\frac{N _{0}}{e} = N_{0} e^{- 3 λ}$

$e^{- 1} = e^{- 3 λ}$

Comparing the powers of $e$ , we get:

$3 λ = 1 \Rightarrow λ = \frac{1}{3} min^{- 1}$

The time in which the activity reduces to half its initial value is the half-life $T_{1/2}$ , which is given by:

$T_{1/2} = \frac{lo g _{e} 2}{λ}$

Substituting the value of $λ$ :

$T_{1/2} = \frac{lo g _{e} 2}{\frac{1}{3}} = 3 lo g_{e} 2 min$

Answer: $3 lo g_{e} 2$

Physics - Modern Physics Question with Solution | TestHub

Options:

Doubts & Discussion