A radioactive element decays by `beta-emission`. A detector records `n` beta particles in `2 s` and in next `2 s` it records `0.75 n` beta particles. Find mean life correct to nearest whole number. Given ln `|2| = 0.6931`, ln `|3|=1.0986`.

Options

A. 7s
B. 9s
C. 11s
D. 15s

Answer

A

Explanation

Suppose `N_(0)` be the initial number of particle. The number of undecayed particles is `N = N_(0) e^(-lambda t)`. Thus number of particles decayed in time t.
`n = N_(0) - N = N_(0) - N_(0) e^(-lambda t) = N_(0) (1 - e^(-lambda t))" "...(i)`
If now N' is the number of particles after t seconds over initial, then undecayed particles 0.75n = N - N'
`= N_(0) e^(-lambda t) - Ne^(-lambda t) = N_(0) e^(-lambda t) - (N_(0) e^(-lambda t))e^(-lambda t)`
`0.75 n = N_(0) e^(-lambda t) (1 - e^(-lambda t))" "...(ii)`
Dividing equation (ii) by (i), we get
`0.75 = e^(-lambda t) or 1n 0.75 = - lambda t`
`therefore" "lambda =- (1n 0.75)/(t) = - (1n (4//3))/(2) = 0.1438 s^(-1)`
Mean life, `tau = (1)/(lambda) = (1)/(0.1438) ~~ 7s`