The surface of a metal is illuminated alternately with photons of energies E₁ = 4eV and E₂ = 2.5eV respectively. The ratio of maximum speeds of the photo electrons emitted in the two cases is 2. The work function of the metal in (eV) is ___________.

From the Einstein’s photoelectric equation Energy of photon
=Kinetic energy of photo electrons +Work function
$\Rightarrow$Kinetic energy =Energy of Photon – Work Function
Let ${\mathrm{\varphi }}_0$ be the work function of metal and v₁ and v₂ be the velocity of photoelectrons. Using Einstein’s photoelectric equation we have
$\begin{array}{l}\frac{1}{2}{\mathrm{mv}}_1^2=4-{\mathrm{\varphi }}_0\dots \dots \dots \dots \dots \dots \dots (\mathrm{i}\text{) }\\ \frac{1}{2}{\mathrm{mv}}_2^2=2.5-{\mathrm{\varphi }}_0\dots \dots \dots \dots \dots \dots \dots \text{ (ii)}\\ \Rightarrow \frac{\frac{1}{2}{\mathrm{mv}}_1^2}{\frac{1}{2}{\mathrm{mv}}_2^2}=\frac{4-{\mathrm{\varphi }}_0}{2.5-{\mathrm{\varphi }}_0}\\ \Rightarrow (2{)}^2=\frac{4-{\mathrm{\varphi }}_0}{2.5-{\mathrm{\varphi }}_0}\\ \Rightarrow 10-4{\mathrm{\varphi }}_0=4-{\mathrm{\varphi }}_0\\ \Rightarrow {\mathrm{\varphi }}_0=2\mathrm{eV}\end{array}$