When a metal surface is illuminated by light of wavelengths 400 nm and 250 nm, the maximum velocities of the photoelectrons ejected are v and 2v respectively. The work function of the metal is - (h = Planck's constant, c = velocity of light in air)
Answer & explanation
Correct answer: option 4
$\frac{1}{2}mv_{max}^2=K.E._{max}=\frac{hc}{λ}-W$
$\frac{1}{2}mv_1^2=\frac{hc}{λ_1}-W$ ….(1)
$\frac{1}{2}mv_2^2=\frac{hc}{λ_2}-W$ ….(2)
$\frac{eq^n(1)}{eq^n(2)}⇒(\frac{v_1}{v_2})^2=\frac{\frac{hc}{λ_1}-W}{\frac{hc}{λ_2}-W}=(\frac{v}{2v})^2$
$\frac{hc}{λ_2}-W=4(\frac{hc}{λ_1}-W)$
$3W=\frac{4hc}{λ_1}-\frac{hc}{λ_2}$
$W=\frac{hc}{3}(\frac{4}{λ_1}-\frac{1}{λ_2})=\frac{hc}{3}[\frac{4}{400×10^{-9}}-\frac{1}{250×10^{-9}}]=\frac{hc}{3}×10^9×\frac{150}{100×250}=0.5hc×10^6J$