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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 of
time t, `(N_(0))/(N ) = e^( lamdat ) , log e ((N_(0))/(N )) = lamda t`
`therefore t = (log e ((N _(0))/(N )))/(lamda )= (2. 303 log _(10) ((N_(0))/(N )))/(lamda)`
In this proble, the initial number of active nuclei, `N_(0) = 6.0 xx 10 ^(18),`
`N = 1.0 xx 10 ^(18) , T = 40s,`
`lamda = (0.693)/(T) = (0.693)/(40) = 1.733 xx 10 ^(-2) s ^(-1).`
`t= (2.303 log _(10) ((6.0 xx 10 ^(18))/( 1. 0 xx 10 ^(18))))/( 1. 733 xx 10 ^(-2))`
`= (2.303 log _(10) (6))/( 1. 733 10 ^(-2))= (2. 303 xx 0.7782)/( 1. 733 xx 10 ^(-2))= 103. 4s.`
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