The surface of a metal is illuminated alternately with photons of...

Given,

Energy of two photons are ${E}_{1} = 4eV$ and ${E}_{2} = 2 . 5eV$

The ratio of maximum speeds of the photoelectrons emitted in the two cases is $\dfrac{{v}_{1}}{{v}_{2}} = 2$

Using Einstein equation of photoelectric effect,

K{E}_\max = \dfrac{1}{2}m{v}^{2} = E - ϕ . . . ( 1 )

Where, $ϕ$ is the work function of metal and $E$ is the energy of photon

Now using equation for both the cases we get,

$\dfrac{1}{2}m{v}_{1}^{2} = 4 - ϕ \ldots ( 2 )$

$\dfrac{1}{2}m{v}_{2}^{2} = 2 . 5 - ϕ . . . ( 3 )$

Dividing equation (2) and (3) and substitute given values, we get,

$\dfrac{{{v}_{1}}^{2}}{{{v}_{2}}^{2}} = \dfrac{4 - ϕ}{2 . 5 - ϕ} = {\left(2\right)}^{2}$

$⇒ 3ϕ = 6$

$⇒ ϕ = 2eV$