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which sample, A or B shown in fig has shorter mean life? |
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Answer» Solution :form fig. , we find that at `t=0, ((dN)/(dt))_(A)=((dN)/(dt))_(B)` Therefore, `(N_(0))_(A)=(N_(0))_(B)` At any subsequent time, `((dN)/(dt))_(B)lt((dN)/(dt))_(A)` `:. lambda_(B)N_(B) lt lambda_(A)N_(A)` As `N_(B) lt N_(A)` (RATE of decay OFB is slower) `:. lambda_(B) lt lambda_(A)` Hence, `tau_(B) lt tau_(A)`, i.e., MEAN life time of B is SHORTER than that of A.
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