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A radioactive element A decays with a de...

A radioactive element `A` decays with a decay constant `lambda`. The fraction of nuclei that decayed at any time `t`, if the initial nucle are `N_(0)` is given by:

A

`e^(-lambda t)`

B

`1 - e^(-lambda t)`

C

`e^(lambda t)`

D

`(1)/(1 - e^(lambda t))`

Text Solution

Verified by Experts

The correct Answer is:
`(b)`
`N = N_(0)e^(-lambda t)`
`:. N_(0) - N = N_(0) - N_(0) e^(-lambda t) = N_(0) [1 - e^(-lambda t)]`
`:. (N_(0) - N)/(N_(0)) = 1 - e^(-lambda t)`
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