**The Photoelectric Effect**

One nature of light is that it acts like a particle. It has been found out that during the emission, absorption and scattering of the electromagnetic radiation, the energy of the wave is *quantized, *meaning it is emitted and absorbed in particle-like packages of definite energy which is called as ** photons**. The

**photoelectric effect**which was explained by Albert Einstein proves this nature of light.

In the photoelectric effect, when a material is illuminated, it emits electrons from its surface. In order for these electrons to escape from the surface of the material, they must absorb enough energy from the light source. After requiring enough energy, the electrons will be able to overcome the attraction of the positive ions in the materials, and escape out of the material.

Einstein made a radical postulate that a beam of light is composed of small packages of energy called photon or quanta. The energy of each photon is equal to a constant, ** h** multiplied by the frequency of that photon,

**. Mathematically, the energy of the photon is expressed as**

*f*\(E=hf\). **(1)**

Since \(f={c \over \lambda}\), **(1)** can be expressed as

\(E={hc \over \lambda}\) **(2)**

where

** c** is the speed of light which is approximately equal to

**3.0 x 10^8 m/s.**

** h** is the Planck’s constant, which is a universal constant, after Max Planck and has a numerical value to the accuracy known at present as \(6.62606957(29) \times 10^{-34}\; J \cdot s\). Mostly, the approximate value usually used in physics problems is \(6.63 \times 10^{-34}\; J \cdot s\).

**Example 1:**

The energy of a photon is 3.0 x 10^{-19} J. What is its frequency?

Solution #1: Since the only given is the energy of a photon, we use the equation \(E = hf\) to solve the problem.

Solution #2: Derive the equation for frequeny, we have \(f={E \over h}\).

Solution #3: Substitue all the given values:

\(f={3.0 \times 10^{-19}\;J \over 6.63 \times 10^{-34}\;J\cdot s}\)

Hint #4: Calculating solution #3, the frequency will be

**\(f=4.5 \times 10^{14}\;Hz\) **

**Example 2:**

A photon of light traveling through space has energy of \(3.3 \times 10^{-19}\;J\). How long must the wave’s length be?

Solution #1: Since the wavelength is what being asked in the problem, we use the equation for energy \(E={hc \over \lambda}\).

Solution #2: Derive the formula for the wavelength, we have \(\lambda = {hc \over E}\).

Solution #3: Substitute the known values:

\(\lambda = {(6.63 \times 10^{-34}\;J \cdot s)(3.0 \times 10^8\;m/s) \over 3.3 \times 10^{-19}\;J}\)

Solution #4: Calculating solution #3, the answer would be

\(\lambda = 6.0 \times 10^{-7}\;m\)

__Figure__

In the figure above, as the light strikes the surface, the process will be all or nothing. That is, the electrons in the material may or may not absorb the photons’ energy. For the electrons to be ejected from the material, they must absorb energy greater than the work function, \(\phi\). **Work function** is the minimum amount of energy needed to eject an electron out of the material. Mathematically, the photoelectrons will only be ejected if \(hf>\phi \; \text{or}\; f>{\phi \over h}\). Therefore, photoelectric effect occurs only for frequencies greater than a minimum threshold frequency.

It is also observed from the postulate that greater photocurrent is caused by greater intensity. This means that, when a particular frequency has greater intensity, a greater number of photons per second is absorbed, thus, a greater number of electrons are emitted per second and so, the photocurrent is also greater.

As soon as a photon with sufficient energy strike the surface of the material, the electrons can absorb them and thus, be ejected out of the material. Therefore, there is no delay between the illumination and emission of photoelectrons.

The postulate also explains that the stopping potential for a given surface depends only on the frequency of light. Applying the law of conservation of energy, the maximum kinetic energy for an emitted electron is the electron’s excess absorbed energy, that is, the energy *hf* gained from a photon minus the work function. Mathematically written as

\(K_{max} = \frac 12 mv_{max}^2 = hf - \phi\). **(3)**

Since \(K_{max} = eV_0\), where \(e\) is the magnitude of electron charge and \(V_0\) is the stopping potential, equation (3) can be expressed as

\(eV_0=hf - \phi\). **(4)**

Equation (4) implies that the stopping potential increases with increasing light frequency.

The electron energy and work function are usually measured in electron volts (eV).

\(1 eV = 1.602 \times 10^{-19} J\)

According to Einstein in his theory of relativity, every particle that has energy must have momentum. A particle with zero rest mass and has an energy, **E** has a momentum, **p** which is given by the equation \(E=pc\). Since photon has zero rest mass, it follows that the momentum of photon is expressed mathematically as

\(p = {E \over c} = {hf \over c} = {h \over \lambda}\).

The direction of the momentum of the photon is the same as the direction of motion of the electromagnetic wave.