# The surface of a metal is illuminated alternately with photons of energies

Question:

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

Solution:

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)

$\Rightarrow \frac{\frac{1}{2} m v_{1}^{2}}{\frac{1}{2} m v_{2}^{2}}=\frac{4-\phi_{0}}{2.5-\phi_{0}}$

$\Rightarrow(2)^{2}=\frac{4-\phi_{0}}{2.5-\phi_{0}} \Rightarrow 10-4 \phi_{0}=4-\phi_{0}$

$\phi_{0}=2 \mathrm{eV}$