QJEEM25Pheb74: JEE Main 2025 Physics - Class 12 JEEM25Pheb74 Solution: In an experiment with photoelectric effect, the stopping…

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.