21.2 Photoelectric Effect

What is the Photoelectric Effect?

The photoelectric effect is the phenomenon in which electrons are emitted from the surface of a metal when electromagnetic radiation of sufficiently high frequency is incident upon it. The emitted electrons are called photoelectrons.

This effect was first observed by Heinrich Hertz in 1887 and later explained by Albert Einstein in 1905 using the concept of photons, for which he received the Nobel Prize in Physics.

Experimental Observations

Key experimental observations that classical wave theory failed to explain:

1. Threshold Frequency: Photoelectrons are only emitted if the frequency of incident light is above a minimum value called the threshold frequency $f_0$. Below $f_0$, no electrons are emitted regardless of how intense the light is.
2. Instantaneous Emission: Electrons are emitted almost instantaneously (within $10^{-9}$ s) once light above $f_0$ strikes the surface — there is no time delay.
3. Independence of K.E. from Intensity: The maximum kinetic energy of emitted photoelectrons depends on the frequency of incident light, not its intensity.
4. Photocurrent and Intensity: The number of photoelectrons emitted per second (photocurrent) is proportional to the intensity of incident light.

Threshold Frequency and Threshold Wavelength

The threshold frequency $f_0$ is the minimum frequency of incident radiation required to eject an electron from a metal surface.

$f_0=\frac{\Phi }{h}$

The corresponding threshold wavelength ${\lambda }_0$ is the maximum wavelength that can cause photoelectric emission:

${\lambda }_0=\frac{c}{f_0}=\frac{hc}{\Phi }$

For $f<f_0$ (or $\lambda >{\lambda }_0$): no emission, regardless of intensity.

Work Function ($\Phi$)

The work function $\Phi$ is the minimum energy required to liberate an electron from the surface of a metal.

$\Phi =h{f}_0$

where $h=6.63\times 10^{-34}$ J s is Planck's constant.

Different metals have different work functions. Metals with lower work functions (e.g., caesium) are more sensitive to light.

Einstein's Photoelectric Equation

Einstein proposed that light consists of discrete packets of energy called photons, each carrying energy $E=hf$.

When a photon strikes a metal surface, it transfers all its energy to a single electron. Part of this energy is used to overcome the work function, and the remainder becomes the kinetic energy of the emitted electron.

By conservation of energy:

$hf=\Phi +K.E_{\max }$

$hf=h{f}_0+\frac{1}{2}m{v}_{\max }^2$

where:

• $h$ = Planck's constant ($6.63\times 10^{-34}$ J s)
• $f$ = frequency of incident light
• $\Phi =h{f}_0$ = work function of the metal
• $K.E_{\max }=\frac{1}{2}m{v}_{\max }^2$ = maximum kinetic energy of emitted electron

Graphical Representation

Rearranging: $K.E_{\max }=hf-\Phi$

This is of the form $y=mx+c$, so a graph of $K.E_{\max }$ vs $f$ is a straight line with:

• Slope = $h$ (Planck's constant)
• x-intercept = $f_0$ (threshold frequency)
• y-intercept = $-\Phi$ (negative of work function)

Stopping Potential ($V_0$)

The stopping potential $V_0$ is the minimum negative potential applied to the anode that just prevents even the most energetic photoelectrons from reaching it, reducing the photocurrent to zero.

$K.E_{\max }=e{V}_0$

where $e=1.6\times 10^{-19}$ C is the charge of an electron.

Combining with Einstein's equation:

$e{V}_0=hf-\Phi$

The stopping potential depends only on the frequency of incident light, not on its intensity.

Why Maximum K.E. is Independent of Intensity

In the photon model, intensity determines the number of photons per second hitting the surface — not the energy of each photon. Since each photoelectron absorbs exactly one photon, the maximum kinetic energy of each electron depends only on the energy of that single photon, i.e., on frequency $f$.

• Increasing intensity → more photons per second → more photoelectrons per second → greater photocurrent
• Increasing intensity → energy per photon unchanged → $K.E_{\max }$ unchanged
• Increasing frequency → each photon has more energy → greater $K.E_{\max }$

This is a fundamental failure of classical wave theory, which predicted that higher intensity should give electrons more energy.

Worked Example

Problem: Light of frequency $8.0\times 10^{14}$ Hz is incident on a metal with work function $\Phi =2.0$ eV. Find the maximum kinetic energy of the emitted photoelectrons.

Solution:

Photon energy: $E=hf=(6.63\times 10^{-34})(8.0\times 10^{14})=5.30\times 10^{-19}$ J $=3.31$ eV

Using Einstein's equation: $K.E_{\max }=hf-\Phi =3.31-2.0=1.31\text{ eV}$