The surface of a metal is illuminated alternately with photons of energies E 1 = 4   eV and E 2 = 2 . 5…

Given,

Energy of two photons are $E_1=4 \mathrm{eV}$ and $E_2=2.5 \mathrm{eV}$

The ratio of maximum speeds of the photoelectrons emitted in the two cases is $\frac{v_1}{v_2}=2$

Using Einstein equation of photoelectric effect,

$K{E}_{max}=\frac{1}{2}m{v}^2=E-\varphi ...\left(1\right)$

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

Now using equation for both the cases we get,

 $\frac{1}{2}m{v}_1^2=4-\varphi \dots \left(2\right)$

$\frac{1}{2}m{v}_2^2=2.5-\varphi ...\left(3\right)$

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

$\frac{{v_1}^2}{{v_2}^2}=\frac{4-\varphi }{2.5-\varphi }={\left(2\right)}^2$

$\Rightarrow 3\varphi =6$

$\Rightarrow \varphi =2 \mathrm{eV}$