The disintegration rate of a certain radioactive sample at any instant is `4750` disintegrations per minute. Five minutes later the rate becomes `2700` per minute. Calculate
(a) decay constant and (b) half-life of the sample

Explanation

We have,
`N =N_(0)e^(-lambda t)`
The activity of the sample is given by
`(dN)/(dt) = - lambda N_(0)e^(- lambda t)= -lambda N`
i.e., activity is proportional to be the number of undercyed nuclei.
At ` t0, ((dN)/(dt))_(t=0) =- lambdan`
At `t=5 min, ((dN).(dt))_(t=5 minm) =- lambdaN`
`:. (N_(0))/(N)=(((dN)/(dt))_(t=0))/(((dN)/(dt))_(t=5min))=(4750)/(2700)=1.760`
a. From Eg. (i), we have
`(N)/(N_(0))=e^(-lambdat)`
or `(N_(0))/(N)=e^(lambdat)`or `log_(e) (N_(0))/(N)=lambda t`
`rArr lambda =(1)/(t) log_(e)(N_(0))/(N)=(2.3026)/(t)xxlog_(10)((N_(0))/(N))`
Substituting give values, we have
`lambda=(2.3026)/(5) xx0.2455=0.113 min^(-1)`
b. Half-life of sample , `T=(0.6931)/(lambda) =(0.6931)/(0.113) =6.1 min^(-1)`.