We know, $$e{V_0} = hv - \phi $$ .... (i)

$$ \Rightarrow e{V_0} = K{E_{\max }}$$ (where $\phi$ = work function of the metal and $v_0$ = stopping potential)

$$ \Rightarrow {V_0} = \left( {{1 \over e}} \right)K{E_{\max }}$$

Stopping potential depends solely on the frequency of the incident light $+(v)$ & the work function of the metal, not the intensity.

From eq. (i), we can see,

$$v \uparrow \Rightarrow {v_0} \uparrow $$

$$ \Rightarrow {e \over {\lambda \downarrow }} \Rightarrow {v_0} \uparrow $$ (as $$v = {c \over \lambda }$$)

So, stopping potential increases with decrease in the wavelength of the incident light.

Hence, option 1 is correct.