Physics Ncert Solutions Class 12th Nuclei Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen Class 12th Physics

Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An example is

³⁸sulphur $\overset{h a l f l i f e}{\to}$ ³⁸Cl $\overset{h a l f l i f e}{\to}$ 38Ar(stable)

Assume that we start 1000 ³⁸S nuclei at time t=0. The number of ³⁸Cl is of count zero at t=0 and will be again zero at t= $\infty$ at what value of t, would the number of counts be a a maximum?

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Alok Kumar Singh | Contributor-Level 10

8 months ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- let ³⁸S have N₁ active nuclei and ³⁸Cl have N₂ active nuclei

$\frac{d N_{1}}{d t} = {- λ}_{1} N_{1}$ + $λ_{1} N_{1}$

N₁=N_{0 $e^{- λ 1 t}$}

$\frac{d N_{2}}{d t} = {- λ}_{1} N_{2} e^{({- λ}_{1} t)}$ + $λ_{2} N_{2}$

Multiplying by $e^{λ 2 t} d t$ and then integrating both sides we got

N2= $\frac{N_{o} λ_{1}}{λ_{2} - λ_{1}} (e^{- λ 1 t} - e^{- λ 2 t})$

After solving it we get time t= (log $\frac{λ_{1}}{λ_{2}}$ )/ $λ_{1} - λ_{2}$

t= $\frac{l o g \frac{2.48}{0.62}}{2.48 - 0.62}$ = $\frac{2.303 * 2 * 0.3010}{1.86} = 0.745 s$

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Physics Ncert Solutions Class 12th 2023

Physics Ncert Solutions Class 12th 2023

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