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A sample of .(19)K^(40) disintegrates in...

A sample of `._(19)K^(40)` disintegrates into two nuclei Ca & Ar with decay constant `lambda _(Ca)=4.5 xx10^(-10) S^(-1)` and `lambda_(Ar)=0.5xx 10^(-10)S^(-1)` respectively. The time after which 99% of `._(19)K^(40)` gets decayed is

A

`6.2xx10^(9) sec`

B

`9.2xx10^(9) sec`

C

`7.2xx10^(9) sec`

D

`4.2xx10^(9) sec`

Text Solution

Verified by Experts

The correct Answer is:
B

t=0
`(dN)/(dt)=-(lambda_(1)+lambda_(2))xxN`
`log_(2) (N/(N_(0)))=(-lambda_(1)+lambda_(2))t`
`2.3xxlog_(10) ((N_(0))/(N_(0)//100))=5xx10^(-10)t`
`(2.303xx2)/(5xx10^(-10))=t`
`2.303xx0.4 xx10^(10) =t`
`t=9.2xx10^(9) sec`
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