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A radioactive substance has 6.0xx10^(18)...

A radioactive substance has `6.0xx10^(18)` active nuclei initially. What time is required for the active nuclei of the same substance to become `1.0xx10^(18)` if its half-life is `40s`.

Text Solution

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The number of active nuclei at any instant to time `t`,
`(N_(0))/(N)=e^(lambdat) , log e((N_(0))/(N))=lambdat`
`:. t=(loge((N_(0))/(N)))/(lambda)=(2.303log_(10)((N_(0))/(N)))/(lambda)`
In this problem, the initial number of active nuclei,
`N_(0)=6.0xx10^(18), N=1.0xx10^(18), T=40s`
` lambda = (0.693)/(T) = (0.693)/(40) = 1.733 xx 10^(-2) s^(-1)`.
`t=(2.303log_(10)((6.0xx10^(18))/(1.0xx10^(18))))/(1.733xx10^(-2))`
`=(2.303log_(10)(6))/(1.733xx10^(2-))=(2.303xx0.7782)/(1.733xx10^(-2))=103.4 s`
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