At time t=0, a material is composed of two ra | vedclass

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Question

${A} \rightarrow {B}, {B} \rightarrow {C}$

$\frac{{d} {N}_{{B}}}{{dt}}=\lambda {N}_{{A}}-\lambda {N}_{{B}}$

$\frac{{d} {N}_{{B}}}{{dt}}=2 \lambd

Vedclass · Accepted Answer

Vedclass · Answer

${A} ightarrow {B}, {B} ightarrow {C}$

$\frac{{d} {N}_{{B}}}{{dt}}=\lambda {N}_{{A}}-\lambda {N}_{{B}}$

$\frac{{d} {N}_{{B}}}{{dt}}=2 \lambda {N}_{{B}_{0}} {e}^{-\lambda t}-\lambda {N}_{{B}}$

${e}^{-\lambda t}\left(\frac{{d} {N}_{{B}}}{{dt}}+\lambda {N}_{{B}}ight)=2 \lambda {N}_{{B}_{0}} {e}^{-\lambda {t}} 	imes {e}^{\lambda {t}}$

$\frac{{d}}{{dt}}\left({N}_{{B}} {e}^{\lambda t}ight)=2 \lambda {N}_{{B}_{0}}$, on integrating

${N}_{{B}} {e}^{\lambda t}=2 \lambda {tN}_{{B}_{0}}+{N}_{{B}_{0}}$

${N}_{{B}}={N}_{{B}_{0}}[1+2 \lambda {t}] {e}^{-\lambda {t}}$

$\frac{{d} {N}_{{B}}}{{dt}}=0$ at $-\lambda[1+2 \lambda {t}) {e}^{-\lambda {t}}+2 \lambda {e}^{-\lambda {t}}=0$

${N}_{{B}_{{max}}}$ at ${t}=\frac{1}{2 \lambda}$

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