Photon Theory Of Light

The photon theory of light, proposed by Albert Einstein, revolutionized our understanding of light by describing it as quantized packets of energy called photons. This theory explains phenomena such as the photoelectric effect, where light ejects electrons from a material, which classical wave theory couldn't account for. In real life, the photon theory of light underpins modern technologies like solar panels, which convert sunlight into electricity and LED lights, which produce light through electron transitions in semiconductors. In this article, we will discuss the concept of Photon Theory, the properties of photons, the mass of photons, the momentum of photons and solved examples for better understanding.

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This Story also Contains

1. Photon Theory of Light
2. Properties of Photon
3. Mass of Photon
4. The Momentum of the Photon
5. Solved Examples Based on Photon Theory of Light
6. Summary

Photon Theory of Light

The photon theory of light, proposed by Albert Einstein in the early 20th century, posits that light is composed of discrete packets of energy known as photons. This theory was a significant departure from the classical wave theory of light, which described light as a continuous wave.

According to Eienstein's quantum theory light propagates in the bundles (packets or quanta) of energy, each bundle being called a photon and possessing energy. The energy of one quantum is given by, hν, where h is Planck's constant and ν is the frequency.

$E=h \nu=\frac{h c}{\lambda}$

$\begin{aligned} & \text { where } \mathrm{c}=\text { Speed of light, } \mathrm{h}=\text { Plank's constant }=6.6 \times 10^{-34} J-\mathrm{sec} \\ & \nu=\text { Frequency in } \mathrm{Hz}, \lambda=\text { Wavelength of light. } \\ & \qquad E(\mathrm{eV})=\frac{12400}{\lambda(Angstrom)}\end{aligned}$

Properties of Photon

1. A photon is a packet of energy (or) particles of light that travels with the speed of light in a straight line.
2. The energy of a photon is given as $E=h \nu$ and it depends on the frequency and it does not change with a change in medium.
3. Photons are electrically neutral and not affected by electric and magnetic fields.
4. Photons do not exist at rest i.e., it is a moving particle
5. The momentum of the photon is given as $p=\frac{h}{\lambda}$
6. All photons of light of a given frequency or wavelength have the same energy or momentum irrespective of light intensity
7. Photons can interact with other particles like electrons, which can be seen in the Compton effect
8. Photons can be created or destroyed when the radiation is emitted or absorbed i.e no if photons are not conserved during the collision
9. The dynamic mass of the photon is m = E/c², where E is the energy of the photon
10. During the photon-electron collision, the momentum and total energy are conserved
11. Photons do not decay on their own
12. The energy possessed by the photon can be transferred to other particles when it interacts with other particles

After energy now let us discuss the mass of the photon.

Mass of Photon

The mass of a photon is a concept that often sparks curiosity in physics. According to our current understanding of quantum mechanics and the theory of relativity, photons are massless particles. They do not have rest mass, which means their mass is zero when they are not moving. However, this does not mean photons lack physical properties; rather, their behaviour and effects are described differently compared to massive particles.

You will study in the theory of relativity that the rest mass of anybody is given by

$
m_v=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}
$

Where: $m_v=$ Relativistic mass (kg)
$
\begin{aligned}
m_o & =\text { Rest mass }(\mathrm{kg}) \\
v & =\text { velocity }\left(\mathrm{ms}^{-1}\right) \\
\mathrm{c} & =\text { speed of light }=3 \times 10^8 \mathrm{~ms}^{-1}
\end{aligned}
$

As the velocity of a photon is the same as the speed of light, so from the above equation we can write that $m_o=0$ . But its effective mass is given as

$E=m c^2=h \nu \Rightarrow m=\frac{E}{c^2}=\frac{h \nu}{c^2}=\frac{h}{c \lambda}$

It is also called as kinetic mass of the photon.

The Momentum of the Photon

The momentum of a photon is a fundamental concept in quantum mechanics and relates to the particle-like behaviour of light. Unlike classical objects, photons, which are particles of light, have momentum despite having no mass. The relationship between the photon's energy and its momentum is described by the equation.

As the momentum of anybody is $=m \cdot v$

Here the velocity = c, i.e., speed of light. So, we can write that

$p=m \times c=\frac{E}{c}=\frac{h \nu}{c}=\frac{h}{\lambda}$

Note

1. In a photon particle collision, total energy and total momentum will be conserved but the number of photons may be changed.

2. All photons of light of a particular freq. (or) wavelengths have the same energy and momentum whatever may be the intensity.

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Solved Examples Based on Photon Theory of Light

Example 1: A charged oil drop is suspended in a uniform field of 3 x 10⁴ V/m so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge = 9.9 × 10^-15 kg and g = 10 m/s²)

1) 3.3 × 10^-18 C

2) 3.2 × 10^-18 C

3) 1.6 × 10^-18 C

4) 4.8 × 10^-18 C

Solution:

Wave-particle duality postulates that all particles exhibit both wave and particle properties

At equilibrium

Net force = 0

$\begin{aligned} & \mathrm{mg}-\mathrm{qE}=0 \Rightarrow q=\frac{\mathrm{mg}}{E} \\ & q=\frac{9.9 \times 10^{-15} \times 10}{3 \times 10^4}=3.3 \times 10^{-18} \mathrm{C}\end{aligned}$

Hence, the answer is the option (1).

Example 2: Microwave oven acts on the principle of :

1) transferring electrons from lower to higher energy levels in the water molecule

2) giving rotational energy to water molecules

3) giving vibrational energy to water molecules

4) giving translational energy to water molecules

Solution:

Microwave energy is absorbed by electrons in a lower energy orbit to get excited into the higher energy orbit in water molecules.

Hence, the answer is the option (1).

Example 3: If a source of power 4 kW produces 10²⁰ photons/second, the radiation belongs to a part of the spectrum called

1) $\gamma$ - rays
2) $X-$ rays
3) ultraviolet rays
4) microwaves

Solution:

The energy of a photon

$E=h \nu=\frac{h c}{\lambda}$

wherein

$h=$ Plank's constant
$\nu=$ frequency of radiation
$\lambda \rightarrow$ wavelength

If the wavelength of the photon is $\lambda$ & n is the number of photons emitted per second, then

$\begin{aligned} & P=n \frac{h c}{\lambda} \\ & 4 \times 10^3=\frac{10^{20} \times 6.62 \times 10^{-34} \times 3 \times 10^8}{\lambda} \\ & \lambda=\frac{19.8 \times 10^{-26} \times 10^{20}}{4 \times 10^3}=4.96 \times 10^{-9} \\ & \lambda=49.6 A\end{aligned}$

This wavelength represents X-rays

Hence, the answer is the option 2.

Example 4: which of the following is the correct relation for the speed of light in a vacuum?

1) $C=\frac{1}{\sqrt{\mu_0 \epsilon_0}}$
2) $C=\sqrt{\mu_0 \epsilon_0}$
3) $C=\frac{1}{\mu_0 \epsilon_0}$
4) $C=\mu_0 \epsilon_0$

Solution:

The velocity of the photon

It travels at the speed of light in a vacuum

which is given as $c=3 \times 10^8 \mathrm{~m} / \mathrm{sec}$

and speed of light in a vacuum is also given as $C=\frac{1}{\sqrt{\mu_0 \epsilon_0}}$

Hence, the answer is the option (1).

Example 5: which of the following is the correct relation for the mass of a photon

1) $m=\frac{h \nu}{c}$
2) $m=\frac{c}{h \nu}$
3) $m=\frac{c^2}{h \nu}$
4) $m=\frac{h \nu}{c^2}$

Solution:

The energy of a photon is given by $E=h \nu$
& energy is also given by $E=m c^2$

So

$m c^2=h \nu \Rightarrow m=\frac{h \nu}{c^2}$

where m=mass of the photon.

We can also see that the photon has zero rest mass

because at $v=0 \Rightarrow m=0$

i.e. Photons can not exist at rest

Hence, the answer is the option (4).

Summary

The photon theory of light, introduced by Albert Einstein, conceptualizes light as discrete packets of energy called photons. This theory explains phenomena such as the photoelectric effect, which classical wave theories could not. Photons are massless particles that travel at the speed of light and exhibit both wave-like and particle-like properties. Despite having no rest mass, photons have energy and momentum, which are fundamental to various technologies, including lasers and photovoltaic cells.