A radioactive nucleus $$X$$ decays to a stable nucleus $$'Y'.$$ Then the graph of rate of formation of $$'Y'$$ against time $$'t'$$ will be

Solution :
$$\eqalign{ & N = {N_0}{e^{ - \lambda t}} \cr & \Rightarrow {N_y} = {N_0}\left( {1 - {e^{ - \lambda t}}} \right) \cr} $$
Rate of formation of $$Y = \frac{{dN}}{{dt}} = + \lambda {N_0}{e^{ - \lambda t}}$$
$$\eqalign{ & X \to Y \cr & {\text{At}}\,\,t = 0,R = \lambda {N_0} \cr & \Rightarrow t = 0\,{N_0} \cr & t = \infty ,R = 0 \cr & \Rightarrow t = t\,N \cr & {N_y} = {N_0}\left( {1 - {e^{ - \lambda t}}} \right) \cr} $$