When two monochromatic lights of frequency, ν and (ν)/(2) are...

Using Einstein's photoelectric equation we can write,

For first case:

$hν = ϕ + e\left(\dfrac{{V}_{S}}{2}\right) . . . \left(1\right)$

For second case:

$h\dfrac{ν}{2} = ϕ + e\left({V}_{S}\right) . . . \left(2\right)$

Using value of $e\left({V}_{S}\right)$ from equation(2) in equation(1), we can write,

$hν = ϕ + \dfrac{1}{2}\left(\dfrac{hν}{2} - ϕ\right)$
$⇒ hν = ϕ + \dfrac{hν}{4} - \dfrac{ϕ}{2}$
$⇒ \dfrac{ϕ}{2} = \dfrac{3hν}{4}$
$⇒ ϕ = \dfrac{3hν}{2}$

As we know, $ϕ = h{ν}_{0}$

Therefore, ${ν}_{0} = \dfrac{3ν}{2}$