The surface of a metal is illuminated alternately with photons of energies $E_1=4 \mathrm{eV}$ and $

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 eV ) is $\_\_\_\_$ .

Professor PC Thomas Classes · Answer

(2) : According to Einstein's photoelectric equation, $$ E-\frac{1}{2} m v^2=\phi $$ For first photon, $$ 4-\frac{1}{2} m v_1^2=\phi \quad 	ext { or } \quad \frac{1}{2} m v_1^2=4-\phi $$ For second photon, $2.5-\frac{1}{2} m v_2^2=\phi$ $$ \frac{1}{2} m v_2^2=2.5-\phi $$ From eq. (i) and (ii), $\left(\frac{v_1}{v_2}ight)^2=\frac{4-\phi}{2.5-\phi}$ $$ 10-4 \phi=4-\phi 	ext { or } 3 \phi=6 \Rightarrow \phi=2.00 \mathrm{eV} $$