Nucleus A decays to B with decay constant λ1 and B decays to C with decay constant λ2. Initially at t = 0, number of nuclei of A and B are 2N0 and N0 respectively. At t = t0, number of nuclei of B stop changing. If at this instant number of nuclei of B are 3N02

dNBdt=λ1NA−λ2NB at, t=t0,dNBdt=0,NB=3N02⇒2N0λ1e−λ1t0−λ03N02=0t0=1λ1ln⁡43λ1λ2NA=λ2NBλ1

Nucleus A decays to B with decay constant λ1 and B decays to C with decay constant λ2. Initially at t = 0, number of nuclei of A and B are 2N0 and N0 respectively. At t = t0, number of nuclei of B stop changing. If at this instant number of nuclei of B are 3N02

dNBdt=λ1NA−λ2NB at, t=t0,dNBdt=0,NB=3N02⇒2N0λ1e−λ1t0−λ03N02=0t0=1λ1ln⁡43λ1λ2NA=λ2NBλ1

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Nucleus A decays to B with decay constant $λ_{1}$ and B decays to C with decay constant $λ_{2}$ . Initially at t = 0, number of nuclei of A and B are 2N₀ and N₀ respectively. At t = t₀, number of nuclei of B stop changing. If at this instant number of nuclei of B are $\frac{3 N_{0}}{2}$

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the value of t₀ is $\frac{1}{λ_{1}} \ln \frac{4}{3} \frac{λ_{1}}{λ_{2}}$

the value of N_A at t₀ is $\frac{3 N_{0}}{2} \frac{λ_{2}}{λ_{1}}$

the value of N_Aat t₀ is $\frac{2 N_{0}}{3} \frac{λ_{2}}{λ_{1}}$

the value of t₀ is $\frac{1}{λ_{2}} \ln \frac{4}{3} \frac{λ_{1}}{λ_{2}}$

answer is A, C.

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Detailed Solution

$\begin{array}{l} \frac{{dN}_{B}}{dt} = λ_{1} N_{A} - λ_{2} N_{B} \\ at, t = t_{0}, \frac{{dN}_{B}}{dt} = 0, N_{B} = \frac{3 N_{0}}{2} \\ \Rightarrow 2 N_{0} λ_{1} e^{- λ_{1} t_{0}} - λ_{0} \frac{3 N_{0}}{2} = 0 \\ t_{0} = \frac{1}{λ_{1}} \ln \frac{4}{3} \frac{λ_{1}}{λ_{2}} \\ N_{A} = \frac{λ_{2} N_{B}}{λ_{1}} \end{array}$