A photocell is illuminated by a small bright source placed 1 m away, number of photo electrons is n. When the same source of light is placed (1/2) m away, the number of electrons emitted by photocathode would be 

     $$\begin{gathered} I=\frac{Pofsource}{4\pi {\left(dis\tan ce\right)}^2}=\frac{P}{4\pi d^2} \Rightarrow e{V}^1=hc[\frac{1}{{\lambda }_3}-\frac{3}{2{\lambda }_1}+\frac{1}{2{\lambda }_2}] \\ \Rightarrow K.E=hc[\frac{1}{{\lambda }_3}-\frac{3}{2{\lambda }_1}+\frac{1}{2{\lambda }_2}] \end{gathered}$$
Here, we assume light to spread uniformly in all directions. Number of photo-electrons emitted from a surface depend on intensity of light / falling on it. Thus the number of electrons emitted n depends directly on I. P remains constant as the source is the same.
$$\begin{gathered} \therefore \frac{I_2}{I_1}=\frac{n_2}{n_1}\Rightarrow \frac{P_2}{P_1}{\left(\frac{d_1}{d_2}\right)}^2=\frac{n_2}{n_1} \\ \therefore \frac{n_2}{n_1}=\left(\frac{P}{P}\right){\left(\frac{1}{1/2}\right)}^2=\frac{4}{1} \Rightarrow n_2=4n[as{n}_1=n] \end{gathered}$$