To determine the half life of a radioactive element , a student plote a graph of in `|(dN(t))/(dt)| versus t , Here |(dN(t))/(dt)|` is the rate of radioatuion decay at time t , if the number of radoactive nuclei of this element decreases by a factor of p after `4.16 ` year the value of p is

Options

A. `2`
B. `4`
C. `6`
D. `8`

Answer

D

Explanation

`N=N_(0) e^(-lambdat)`
`(dN)/(dt)= -lambda N_(0) e^(-lambdat)`
`|(dN)/(dt)|=lambdaN_(0)e^(-lambdat)`
`log|(dN)/(dt)|=log_(e) lambdaN_(0)-lambdat`
`log|(dN)/(dt)|=-lambdat+log_(e) (lambdaN_(0))...........(i)`
Comparing if with the equation of straight line
`y=mx+C`, we get
Slope`=-lambda=(3-4)/(6-4)` (form graph)
or `lambda=1/2yr^(-1)`
Half life, `T=0.6931/lambda=2xx0.6931=1.386yr`
`N=N_(0).(1/2)^(t//T)` or `N/(N_(0))=(1/2)^(t//T)`
`:. 1/p=(1/2)^(4.16/1.386)=(1/2)^(3)=1/8` or `p=8`