Correct Answer - C
`N=N_(0)e^(-lambdat)` से,
`(N)/(N_(0))e^(-lambdat)implies(2)/(3)e^(-lambdat_(1))` तथा `=e^(-lambdat_(2))`
`:.(2)/(3)xx(3)/(1)=(e^(-lambdat_(1)))/(e^(-lambdat_(2)))implies2=e^(lambda(t_(2)-t_(1)))`
`implieslog_(e)2=lambda(t_(2)-t_(1))`
`implies(log_(e)2)/(lambda)=t_(2)-t_(1)`, किंतु `T_(1//2)(log_(e)2)/(lambda)`
`:. t_(2)-t_(1)=T_(1//2)=20` मिनट |