The frequency and the intensity of incident beam of light falling...

Using Einstein's photoelectric equations,

$hv = ϕ + e{V}_{s}$

$e{V}_{s} = hv - ϕ.... ( i )$

Thus, when the frequency of incident light radiation (photons) $v$ increases to $2v$, the stopping potential ${V}_{s}$ changes to ${V}_{s}$

$e{V}_s^{'} = h ( 2v ) - ϕ.... \left(ii\right)$

On dividing Eq. (ii) by Eq. (ii), we get

$\dfrac{{V}_{s} '}{Vs} = \dfrac{2hv - ϕ}{hv - ϕ} = \dfrac{2hv - 2ϕ + ϕ}{hv - ϕ}$

$= \dfrac{2 ( hv - ϕ ) + ϕ}{hv - ϕ} =2 + \dfrac{ϕ}{hv - ϕ}$

$∴ \dfrac{V^{'}s}{Vs}$> 2 ⇒ {V}_s^{'} > 2{V}_{s}$</p>$