1.

A radioactive sample emit n beta - particles is 2 sec , In next 2 sec it emits 0.75 n beta- particle , what is the mean life of the sample?

Answer»


Solution :LET initial number of nuclei `=N_(0)`
`N_(0)overset(2 "sec")rarr (N_(0)-N) overset(2 "sec")rarr(N_(0)-n-0.75 n)`
(No. of `beta` particles emitted=No. of nuclei disintegrated)
No. of nuclei disintegrated in time is given by `=N_(0)(1-e^(-lambdat t)`
`:.` for initial 2 seconds
`n=N_(0)(1-e^(-2lambda)) "" `...........(i)
for next2 seconds
`0.75 n=(N_(0)-n)(1-e^(-2 lambda)) "" ` ..........(ii)
Subtracting (ii) from (i) ((i)-(ii))
`.25 n=n (1-e^(-2lambda))`
`e^(-2lambda)=0.75`
`-2lambda=ln. (3)/(4) "" rArr 2lambda =ln .(4)/(3)`
`t_("avg")=(1)/(lambda)=(2)/(ln((4)/(3)))=6.95 "sec"`


Discussion

No Comment Found

Related InterviewSolutions