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Photoelectric Effect and WaveParticle Duality — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsPhotoelectric Effect and WaveParticle Duality5 Marks
Question
A horizontal cesium plate ($\phi$ = 1.9eV) is moved vertically downward at a constant speed u in a room full of radiation of wavelength 250 run and above. What should be the minimum value of u so that the vertically upward component of velocity is nonpositive for each photoelectron?
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Answer

When $\lambda=250\text{nm}$Energy of photon
$=\frac{\text{hc}}{\lambda}=\frac{1240}{250}=4.96\text{ev}$
$\therefore\text{K.E.}=\frac{\text{hc}}{\lambda}-\text{w}=4.96-1.9\text{ev.}$
Velocity to be non positive for each photo electron
The minimum value of velocity of plate should be = velocity of photo electron
$\therefore$ Velocity of photo electron $=\sqrt{\frac{2\text{KE}}{\text{m}}}$
$=\sqrt{\frac{3.06}{9.1\times10^{-31}}}$
$=\sqrt{\frac{3.06\times1.6\times10^{-19}}{9.1\times10^{-31}}}=1.04\times10^6\text{m/sec.}$

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