A) 1
B) 2
C) 5
D) 4
Correct Answer: B
Solution :
[b] \[\frac{d{{N}_{A}}}{dt}=-{{\lambda }_{1}}{{N}_{A}},\,\,\,\frac{d{{N}_{B}}}{dt}=2{{\lambda }_{1}}{{N}_{A}}-{{\lambda }_{2}}{{N}_{B}},\] \[{{N}_{B}}=\] maximum \[\Rightarrow \,\,\frac{d{{N}_{B}}}{dt}=0\] \[\Rightarrow \,\,\,\,\,\,\,\,2{{\lambda }_{1}}{{N}_{A}}={{\lambda }_{2}}{{N}_{{{B}_{\max }}}}\Rightarrow {{N}_{{{B}_{\max }}}}=\frac{2{{\lambda }_{1}}}{{{\lambda }_{2}}}{{N}_{A}}\] \[\Rightarrow \,\,\,\,\,\,\,\,{{N}_{{{B}_{\max }}}}=\frac{2{{\lambda }_{1}}}{{{\lambda }_{2}}}{{N}_{0}}{{e}^{-{{\lambda }_{{{1}^{t}}=2}}}}\]You need to login to perform this action.
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