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State the law of radioactivity and hence, show thatN=N_(0)e^(-lambda t). |
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Answer» Solution :(a) `(dN)/( DT) = - lambda N` `int_(N_0)^(N) (dN)/( N) = int_(0)^(t) -lambda dt` `[log_(E)^(N) ]_(N_0)^(N) = - lambda [t]_(0)^(t)` `log_(e) (N)/(N_0)=- lambda t` `N=N_(0) e^(- lambda t)` (b) (i) `""{:(22),(11):} Na to {:(22),(11):} Ne + e^(+) +v` or, `{:(22),(11):} Na to {:(22),(11):} Xe + e^(+) + v` Basic PROCESS `p to n + e^(+) + v` (II) Isobar. |
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