State law of radioactive decay. Hence derive the relation N = N _ o e ^ - t . Represent it graphically.

Radioactive decay law : The number of nuclei undergoing the decay per unit time is proportional to the number of unchanged nuclei present at that instant. Proof: * Let ‘N’ be the number of nuclei present at any instant ‘t’. * Let 'dN’ be the number of nucleic that disintegrated in short interval of time ‘dt’. * According to law - dN dt N dN dt =- N ...........(1) Where λ - decayconst dN N =- dt integrating both sides dN N = - dt (e l l) N = t+C ^ N ^ = t + C ^ ..........(2) Where C integration consta at t = 0 , N =N_0 nN _0=C ∴ Frome equation (2) and (3) N=- t+ n N_0 t=In N-In N_0 - T = (N N_0 ) N N_0 =e^ -( t) N=N_0 e^ -( t)

State law of radioactive decay. Hence derive the relation N = N _…

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Question

State law of radioactive decay. Hence derive the relation $N = N _{ o } e ^{-\lambda t}$. Represent it graphically.

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Answer

Radioactive decay law :
The number of nuclei undergoing the decay per unit time is proportional to the number of unchanged nuclei present at that instant.
Proof:

Let ‘N’ be the number of nuclei present at any instant ‘t’.
Let 'dN’ be the number of nucleic that disintegrated in short interval of time ‘dt’.
According to law

$\frac{- dN }{ dt } \propto N$
$\therefore \frac{ dN }{ dt }=-\lambda N ...........(1)$
Where λ - decayconst
$\therefore \frac{ dN }{ N }=-\lambda dt$
$\therefore$ integrating both sides
$\int \frac{ dN }{ N }=\int-\lambda dt$
${(e l l) N =\lambda t+C \ell^{\prime \prime} N ^{\prime \prime}=\lambda t + C ^{\prime}}..........(2)$
Where $C \rightarrow$ integration consta

$\therefore at t = 0 , N =N_0$
$\therefore \ell nN _0=C$
∴ Frome equation (2) and (3)
$\ln N=-\lambda t+\ell n N_0$
$\lambda t=\operatorname{In} N-\operatorname{In} N_0$
$-\lambda T =\ln \left(\frac{N}{N_0}\right)$
$\frac{N}{N_0}=e^{-(\lambda t)}$
$\therefore N=N_0 e^{-(\lambda t)}$