${N}_{{A}}(0)=$ ${t}=0$ સમયે $A$ ના પરમાણુ
${N}_{{B}}(0)=$ ${t}=0$ સમયે $B$ ના પરમાણુ
\(\frac{{d} {N}_{{B}}}{{dt}}=\lambda {N}_{{A}}-\lambda {N}_{{B}}\)
\(\frac{{d} {N}_{{B}}}{{dt}}=2 \lambda {N}_{{B}_{0}} {e}^{-\lambda t}-\lambda {N}_{{B}}\)
\({e}^{-\lambda t}\left(\frac{{d} {N}_{{B}}}{{dt}}+\lambda {N}_{{B}}\right)=2 \lambda {N}_{{B}_{0}} {e}^{-\lambda {t}} \times {e}^{\lambda {t}}\)
\(\frac{{d}}{{dt}}\left({N}_{{B}} {e}^{\lambda t}\right)=2 \lambda {N}_{{B}_{0}}\), 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 \({t}=\frac{1}{2 \lambda}\)
$4\,{\,_1}{H^1}\, \to \,{\,_2}H{e^4} + 2\,{\,_1}{e^0}\, + \,\,2\,v\,\, + 26\,\,MeV\,\,$