To determine the half life of a radioact...

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

Text Solution

Verified by Experts

The correct Answer is:

`N =N_0 e^(-lambda t)`
`ln|dN|dt|=1n(N_0 lambda)-lambda t`
From graph
`lambda =(1)/(2)` per year
`(t_(1))/(2)=(0.693)/(1//2)=1.386` year
`4.16` years `=3t_(1//2)`
`p=8`.

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12 half life (T_12) of radioactive element is related to its average life (T_(aw)) as

`T_12=1.44 T_(av)`

`T_12=1.36 T_(av)`

`T_12=0.693 T_(av)`

`T_12= (T_(av))/0.693`

The half life of a radioactive element is T and its initial activity at t=0 is A_(0) and at t=t it is A, then

`A=A_(0) 2^(t//T)`

`A=A_(0)(2t)^(T)`

`A_(0)=A 2^(t//T)`

`A_(0)=A 2^(-t//T)`

The half life of a radioactive substance is T_(0) . At t=0 ,the number of active nuclei are N_(0) . Select the correct alternative.

The number of nuclei decayed in time internal `0-t` is `N_(0)e^(-lambda t)`

The number of nuclei decays in time interval `0-t` is `N_(0) (1-e^(-lambda t))`

The probability that a radioactive nuclei does not decay in interval `0-t` is `e^(-lambdat)`

The probability that a radioactive nuclei does not decay in interval `1-e^(-lambdat)`