In an experiment with photoelectric effect, the stopping potential,

<p>We know, $e{V_0} = hv - \phi$ .... (i)</p> <p>$\Rightarrow e{V_0} = K{E_{\max }}$ (where $\phi$ = work function of the metal and $v_0$ = stopping potential)</p> <p>$\Rightarrow {V_0} = \left( {{1 \over e}} \right)K{E_{\max }}$</p> <p>Stopping potential depends solely on the frequency of the incident light $+(v)$ & the work function of the metal, not the intensity.</p> <p>From eq. (i), we can see,</p> <p>$v \uparrow \Rightarrow {v_0} \uparrow$</p> <p>$\Rightarrow {e \over {\lambda \downarrow }} \Rightarrow {v_0} \uparrow$ (as $v = {c \over \lambda }$)</p> <p>So, stopping potential increases with decrease in the wavelength of the incident light.</p> <p>Hence, option 1 is correct.</p>