The surface of a metal is illuminated alternately with photons of energies $E_{1}=4 \mathrm{eV}$ and $E_{2}=2.5 \mathrm{eV}$ respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is 2 . The work function of the metal in $(\mathrm{eV})$ is
From the Einstein's photoelectric equation
Energy of photon
$=$ Kinetic energy of photoelectrons + Work function
$\Rightarrow$ Kinetic energy $=$ Energy of Photon $-$ Work Function
Let $\phi_{0}$ be the work function of metal and $v_{1}$ and $v_{2}$ be the velocity of photoelectrons. Using Einstein's photoelectric equation we have
$\frac{1}{2} m v_{1}^{2}=4-\phi_{0}$ ...(1)
$\frac{1}{2} m v_{2}^{2}=2.5-\phi_{0}$ ...(2)
$\Rightarrow \frac{\frac{1}{2} m v_{1}^{2}}{\frac{1}{2} m v_{2}^{2}}=\frac{4-\phi_{0}}{2.5-\phi_{0}}$
$\Rightarrow(2)^{2}=\frac{4-\phi_{0}}{2.5-\phi_{0}} \Rightarrow 10-4 \phi_{0}=4-\phi_{0}$
$\phi_{0}=2 \mathrm{eV}$