InterviewSolution
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InterviewSolution
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Let `R_(t)` represents activity of a sample at an insant and `N_(t)` represent number of active nuclei in the sample at the instant. `T_(1//2)` represents the half life. `{:(,"Column I",,"Column II"),((A),t=T_(1//2),(p),R_(t)=(R_(0))/(2)),((B),t=(T_(1//2))/(ln2),(q),N_(0)-N_(t)=(N_(0))/(2)),((C),t=(3)/(2)T_(1//2),(r),(R_(t)-R_(0))/(R_(0)) = (1-e)/(e)),(,,(s),N_(t)=(N_(0))/(2sqrt(2))):}` |
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Answer» Correct Answer - (A) p,r (B) r (C) s (A) In half life active sample reduce `=R_(0)/2` `:.` Decay number of nuclei is `=R_(0)/2` (B) `N=N_(0)e^(-lambdat)` where `lambda=` decay constant, `lambda=(ln(2))/t_(1//2)` `N=N_(0)e^(-(ln2t_(1//2))/(t_(1//2)ln2))implies N=N_(0)/e` `N_(0)-N=N_(0) [(e-1)/e]implies (N-N_(0))/N_(0) =(1-e)/e` (C) `N=N_(0)/((2)^(t//T_(1//2)))implies N=N_(0)/2^(3//2)implies N_(0)/(2sqrt(2))` |
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