Question

A horizontal cesium plate ($\Phi$ = 1.9 eV) is moved vertically downward at a constant speed v in a room full of radiation of wavelength 250 nm and above. What should be the minimum value of v so that the vertically upward component of velocity is nonpositive for each photo electron ?

Answer 1

When $\lambda$ = 250 nm

Energy of Photon

$\frac{hc}{\lambda }=\frac{1240}{250}=4.96eV$

$\therefore \text{K.E.}=\frac{hc}{\lambda }-W=4.96-1.9eV$

$=3.06eV$

Velocity to be non positive for each photo electron

The minimum value of velocity of plate should be equal to velocity of photo electron

$\therefore$ Velocity of photo electron

$=\sqrt{\frac{2KE}{M}}=\sqrt{\frac{2\times 3.06\times 1.6\times {10}^{-19}}{9.1\times {10}^{-31}}}$

$=1.04\times {10}^6\text{m/sec}$