JEE Physics Dual Nature Of Radiation and Matter Notes Set 04

Photoelectric Effect

When light of suitable frequency falls on a metal surface, electrons are emitted. Experiments show that emission occurs only if the frequency exceeds a threshold \( f_0 \) that depends on the metal, that emission is almost instantaneous, and that the maximum kinetic energy rises with frequency but does not depend on intensity.
Einstein explained this by treating light as photons of energy \( E=hf \). One photon gives all its energy to one electron: \( hf=\phi+K_{max} \) with work function \( \phi=hf_0 \). The stopping potential satisfies \( eV_s=K_{max} \).
Intensity decides how many photons arrive per second, so it controls the number of photoelectrons and the saturation current only. Wave theory cannot explain the threshold or the delay-free emission.

Photon Momentum and Radiation Pressure

A photon has energy \( E=hf=\frac{hc}{\lambda} \) and momentum \( p=\frac{h}{\lambda}=\frac{E}{c} \). Light that is fully absorbed delivers momentum E/c; light that is reflected straight back delivers 2E/c.
A handy conversion is \( E(\text{eV})=\frac{1240}{\lambda(\text{nm})} \).

de Broglie Waves

Every moving particle has an associated wavelength \( \lambda=\frac{h}{p}=\frac{h}{mv} \). For a particle of kinetic energy K, \( \lambda=\frac{h}{\sqrt{2mK}} \).
A charge q accelerated through potential V has \( \lambda=\frac{h}{\sqrt{2mqV}} \). For electrons this is \( \lambda\approx\frac{12.27}{\sqrt V}\ \text{Å} \). So the wavelength falls as \( \frac{1}{\sqrt V} \), and for equal kinetic energy lighter particles have longer wavelengths.