For radioactive disintegration of a radioactive substance, show that
`N = N_(0)e^(-lambdat)`
where the terms have their usual meaning,

Explanation

(a) The following laws, known as the laws of radioactive decay:
(1) It is a spontaneous phenomenon and one cannot predict, when a particular atom will undergo disintergration. According to radioactive decay law,
`(dN)/(dt)propN " "(dN)/(dt)=lambdaN`.....(i)
From the equation (i)
`(dN)/(dt)= -lambdadt`
Intergrating, we have
`int (dN)/(N)= -lambdaint dt`
or `Log_(e )N= -lambdat+K`, ......(ii)
Where `K` is constant of intergration,
when `t=0, N=N_(0)`
On Setting `t=0` and `N=N_(0)`
the equation (ii)
`log_(e ) N_(0)= -lambdaxx0+k`
`k=log_(e )N_(0)`
Substituting for `K` in the equation (ii)
`log_(e )N= -lambdat+log_(e )N_(0)`
`"log"_(e )(N)/(N_(0))-lambda t , (N)/(N_(0))-e^(-lambdat)`
`N=N_(0)e^(-lambda t)`
(b) (i) The fission reaction of `._(92)uu^(325)` may be represented as given below:
`{:(._(92).^(235),._(0)n^(1),[._(92).^(236)],),(._(56)Ba^(141),._(36)Kr^(92),3_(0)n^(1),Q):}`
The energy `(Q)` released was estimated to be `200 MeV` per fission (or about `0.9 MeV` per nucleon) and is equivalent to the diffecence in masses of the nuclei before and after the fission.
(ii) When two or more than two light nuclei fuse together to form harry nucleus wiht the liberation of energy, the releasing `24 MeV` of energy. The fusion reaction may be expressed as follow:
` H^(2)+_(1)H^(2)rarr_(2)H^(4)e+24MeV`
The above nuclear fusion reaction is energetically possible, only if the mass of the `_(2)H^(4)e` nucleus is less than the sum of the massess of wo deuteron nuclei.