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Derive the relation N_(t) =N_(0)e^(-lambdat) for radioactive decay. Obtain the relation between disintegration constant and half-life. |
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Answer» SOLUTION :According to the LAW of radioactive decay, we know that rate of disintegration of a radioactive sample is directly PROPORTIONAL to the actual quantity of that material at that instant. Mathematically, `-(dN)/(DT)=lambdaN` where `lambda` is the disintegration (or decay) constant of given radioactive substance `therefore (dN)/(dt)=lambdadt` On integration, we have `int_(N_(0))^(N_(t))(dN)/N=-lambdaint_(0)^(t)dt implies [log_(e)N]_(N_(0))^(N_(t))=-lambda[t]_(0)^(t) implies log_(e)N_(t)-log_(e)N_(0)=-lambdat " or" log_(e)(N_(t))/(N_(0))=-lambdat " or" N_(t)/N_(0)=e^(-lambdat) implies N_(t)=N_(0)e^(-lambdat)` For relation between `T_(1/2)` and `lambda`, see Short Answer QUESTION Number 14. |
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