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The decay constant of a radioactive isot...

The decay constant of a radioactive isotope is `lambda`. If `A_1` and `A_2` are its activities at time `t_1` and `t_2` respectively, then the number of nuclei which have decayed the time `(t_1-t_2)`

A

`A_1t_1-A_2t_2`

B

`A_1-A_2`

C

`(A_1-A_2)//lambda`

D

`lambda(A_1-A_2)`

Text Solution

Verified by Experts

The correct Answer is:
C
`A_1=lambdaN_1` at time `t_1`
`A_2=lambdaN_2` at time `t_2`
Therefore, number of nuclei decayed during time interval `(t_1-t_2)` is `N_1-N_2=((A_1-A_2))/lambda`
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