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A radioactiev nuleus X deays to a stable...

A radioactiev nuleus `X` deays to a stable nuleus `Y`. Then, time graph of rate of formation of `Y` against time `t` will be:

A

B

C

D

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The correct Answer is:
c
`N=N_(0) e^(- lmabda t),N_(Y) =N_(0)(1-e^(-lambda t))`
Rate of formation of `Y` is `(dN)/(dt) =+ lambda N_(0)e^(-lambda t)` Which decreases exponentially with time.
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