JEE Physics Dual Nature Of Radiation and Matter Notes Set 03

Photoelectric Effect: Observations and Einstein's Explanation

When light of suitable frequency falls on a metal surface, electrons are emitted almost instantly. The emission starts only if the frequency exceeds a threshold value \( f_0 \) that depends on the metal, whatever the intensity. Below \( f_0 \) no electrons come out, even for very intense light.
For a fixed frequency, the photocurrent is proportional to the intensity (number of photons per second), while the maximum kinetic energy of the electrons does not depend on intensity. The maximum kinetic energy rises linearly with frequency.
These facts cannot be explained by the wave theory, so light is treated as a stream of photons, each of energy \( E=hf=\frac{hc}{\lambda} \). One photon is absorbed by one electron. A part \( \phi=hf_0 \) (the work function) is spent in leaving the metal and the rest appears as kinetic energy:
\[ hf=\phi+K_{max}\qquad\Rightarrow\qquad K_{max}=hf-\phi \]
Useful numbers: \( hc=12400\ \text{eV Å}=1240\ \text{eV nm} \), and \( 1\ \text{eV}=1.6\times10^{-19}\ \text{J} \).

Stopping Potential and Graphs

The retarding potential at which even the fastest photoelectron is just stopped is the stopping potential \( V_0 \): \( eV_0=K_{max}=hf-\phi \). It depends on frequency and the metal only. A graph of \( V_0 \) against f is a straight line of slope \( h/e \) with intercept \( f_0 \) on the frequency axis; the same slope is obtained for every metal.
The current against anode potential graph rises to a saturation value, which is proportional to intensity. All curves for the same frequency cut the voltage axis at the same \( -V_0 \).

Photons, Momentum and Radiation Pressure

A photon has zero rest mass, energy \( E=pc \) and momentum \( p=\frac{h}{\lambda}=\frac{E}{c} \).
For a beam of intensity I falling normally on a surface, the pressure is \( \frac{I}{c} \) if it is completely absorbed and \( \frac{2I}{c} \) if it is completely reflected. For partial reflection with coefficient \( \rho \) at incidence angle \( \theta \), the normal pressure is \( \frac{I}{c}(1+\rho)\cos^2\theta \). The force on an area A is pressure times A (the cos factor accounts for the oblique spread of the beam).
Number of photons per second in a beam of power P: \( n=\frac{P}{hf} \). The photocurrent is \( I=n\eta e \), where \( \eta \) is the fraction of photons that eject an electron.

Matter Waves and the Davisson-Germer Experiment

De Broglie suggested that every moving particle has a wavelength \( \lambda=\frac{h}{p}=\frac{h}{mv} \). For kinetic energy K it is \( \lambda=\frac{h}{\sqrt{2mK}} \).
For a charge q accelerated through a potential difference V: \( \lambda=\frac{h}{\sqrt{2mqV}} \). For an electron this gives \( \lambda=\frac{12.27}{\sqrt V}\ \text{Å} \); for a proton \( \lambda=\frac{0.286}{\sqrt V}\ \text{Å} \); for an alpha particle \( \lambda=\frac{0.101}{\sqrt V}\ \text{Å} \).
For a particle in thermal equilibrium, \( \lambda=\frac{h}{\sqrt{3mkT}} \).
The wave nature of electrons was confirmed by Davisson and Germer, who scattered electrons from a nickel crystal and found a strong peak at a particular angle for 54 V electrons, in agreement with Bragg's law \( 2d\sin\theta=n\lambda \) and with the de Broglie wavelength of about 1.67 Å.

Worked Examples

Worked Example 1. Light of wavelength 400 nm falls on a metal of work function 2.0 eV. Find the stopping potential and the threshold wavelength.
Answer:
Photon energy \( =\frac{1240}{400}=3.1\ \text{eV} \), so \( eV_0=3.1-2.0=1.1\ \text{eV} \) and \( V_0=1.1\ \text{V} \).
Threshold wavelength \( \lambda_0=\frac{1240}{2.0}=620\ \text{nm} \).

Worked Example 2. A 100 W lamp emits 5% of its power as light of wavelength 600 nm. How many photons does it emit per second?
Answer:
Light power \( =5\ \text{W} \). Energy of a photon \( =\frac{hc}{\lambda}=\frac{6.63\times10^{-34}\times3\times10^8}{6\times10^{-7}}=3.3\times10^{-19}\ \text{J} \).
Photons per second \( =\frac{5}{3.3\times10^{-19}}\approx1.5\times10^{19} \).

Worked Example 3. Find the de Broglie wavelength of an electron accelerated from rest through 150 V.
Answer:
\( \lambda=\frac{12.27}{\sqrt{150}}=1.0\ \text{Å} \).

Worked Example 4. A beam of light of intensity 3 x 10⁴ W/m² falls normally on a perfectly absorbing plate of area 2 cm². Find the force on the plate.
Answer:
Pressure \( =\frac{I}{c}=\frac{3\times10^4}{3\times10^8}=10^{-4}\ \text{N/m}^2 \).
Force \( =10^{-4}\times2\times10^{-4}=2\times10^{-8}\ \text{N} \).