A)
B)
C)
D)
Correct Answer: C
Solution :
[c] \[N={{N}_{0}}{{e}^{-\lambda t}}\Rightarrow {{N}_{y}}={{N}_{0}}(1-{{e}^{-\lambda t}})\] Rate of formation of \[Y=\frac{dN}{dt}=+\lambda {{N}_{0}}{{e}^{-\lambda t}}\] \[X\xrightarrow{{}}Y\] At \[t=0,\,\,\,\,\,\,\,R=\lambda {{N}_{0}}\Rightarrow t=0\,\,\,{{N}_{0}}\] \[t=\infty ,\,\,\,R=0\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,t=t\,\,\,\,\,\,\,\,\,N\] \[{{N}_{y}}={{N}_{0}}(1-{{e}^{-\lambda t}})\]
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