A metal exposed to light of wavelength 800nm and emits...

According to Einstein's photoelectric equation

{k}_\max = \dfrac{hc}{\lambda} - ϕ

$⇒ {k}_{1} = \dfrac{1230}{800} - ϕ . . . \left(1\right)$

And $⇒ {k}_{2} = \dfrac{1230}{500} - ϕ$

Since, ${k}_{2} = 2{k}_{1}$

$⇒ 2{k}_{1} = \dfrac{1230}{500} - ϕ . . . \left(2\right)$

Eliminating ${k}_{1}$ from (1) and (2) we get

$0 = \dfrac{1230}{500} - \dfrac{1230}{400} + ϕ$

$⇒ ϕ = 0 . 615eV$