1.

which sample, A or B shown in fig has shorter mean life?

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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