The study of Physics Topics involves the exploration of matter, energy, and the forces that govern the universe.
What are the Important Properties of Photons?
Photon: It has already been stated that photoelectric effect cannot be explained in terms of wave theory of light. In 1905, Einstein used Planck’s quantum theory and introduced the concept of photon particles. Thus, he could explain photoelectric effect. Particle concept of radiation is the basis of quantum theory. Basic point of the theory is:
Electromagnetic radiation is not wave by nature but consists of a stream of particles called photons.
Properties of Photons: Main properties of photons are-
- Photons are electrically neutral.
- Photons travel with the speed of light, which does not change under any circumstances. (velocity of light, c = 3 × 108 m ᐧ s-1)
- Energy carried by a photon, E = hf; where, f = frequency of radiation and h = Planck’s constant. The energy radiated increases with the increase in number of photons in its stream and hence the intensity of radiation also increases.
- According to Einstein’s theory of relativity, rest mass of a particle is zero, if it travels at the speed of light. Hence, rest mass of each photon is zero.
- By the theory of relativity, if the rest mass of a particle is m0 and its momentum is p, energy of the particle,
E = \(\sqrt{p^2 c^2+m_0^2 c^4}\). In case of a photon, m0 = 0, hence E = pc, or p = \(\frac{E}{c}\) = hf/c. Thus, in spite of photon being a massless particle, it has a definite momentum.
Planck’s constant: It is a universal constant.
unit of h in SI \(=\frac{\text { unit of } E}{\text { unit of } f}\) = \(\frac{\mathrm{J}}{\mathrm{s}^{-1}}\) = J ᐧ s
Unit of h in CGS system = erg ᐧ s.
This unit is same as the unit of angular momentum. Thus h is actually a measure of angular momentum.
Value of h = 6.625 × 10-34 J ᐧ s = 6.625 × 10-27 erg ᐧ s
Relation between the wavelength of radiation and the photon energy: Energy of a photon, E = hf = \(\frac{h c}{\lambda}\)
1 eV = 1.6 × 10-19J and 1Å = 10-10 m;
Hence, expressing E in eV unit and λ in Å unit
λÅ = λ × 10-10m E eV = E × (1.6 × 10-19)
Hence, E × (1.6 × 10-19) = \(\frac{\left(6.625 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{\lambda \times 10^{-10}}\)
or, E = \(\frac{\left(6.625 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{\left(1.6 \times 10^{-19}\right) \times \lambda \times 10^{-10}}\) ≈ \(\frac{12422}{\lambda}\)
Usually, the number on the right hand side is taken as 12400
Hence,
Numerical Examples
Example 1.
Find the energy of a photon of wavelength 4950 Å in eV (h = 6.62 × 10-27 erg ᐧ s). What is the momentum of this photon?
Solution:
λ = 4950Å = 4950 × 10-8 cm
Hence energy of a photon,
E = hf = \(\frac{h c}{\lambda}\) = \(\frac{6.62 \times 10^{-27} \times 3 \times 10^{10}}{4950 \times 10^{-8}}\)
= 4.012 × 10-12 erg
= \(\frac{4.012 \times 10^{-12}}{1.6 \times 10^{-12}}\) = 2.5 eV
Momentum of photon,
p = \(\frac{E}{c}\) = \(\frac{4.012 \times 10^{-12}}{3 \times 10^{10}}\)
= 1.34 × 10-22dyn ᐧ s
Momentum of photon,
p = \(\frac{E}{c}\) = \(\frac{4.012 \times 10^{-12}}{3 \times 10^{10}}\)
= 1.34 × 10-22dyn ᐧ s
Example 2.
Find the number of photons emitted per second by a source of power 25W. Assume, wavelength of emitted light = 6000Å. h = 6.62 × 10-34 J ᐧ s.
Solution:
λ = 6000 Å = 6000 × 10-10 m, c = 3 × 108 m ᐧ s-1
Number of photons per second emitted by a source of power P is,
n = \(\frac{P}{h f}\) = \(\frac{P}{h c / \lambda}\) = \(\frac{P \lambda}{h c}\)
∴ n = \(\frac{25 \times 6000 \times 10^{-10}}{6.62 \times 10^{-34} \times 3 \times 10^8}\)
= 7.55 × 1019
Example 3.
Wavelength of ultraviolet light is 3 × 10-5 cm. What will be the energy of a photon of this light, in eV? (c = 3 × 1010m ᐧ s-1) [HS’06]
Solution:
Planck’s constant, h = 6.625 × 10-27 erg ᐧ s
Wavelength, λ = 3 × 10-5 cm
∴ Energy of a photon,
E = \(\frac{h c}{\lambda}\) = \(\frac{\left(6.625 \times 10^{-27}\right) \times\left(3 \times 10^{10}\right)}{3 \times 10^{-5}}\)erg
= \(\frac{\left(6.625 \times 10^{-27}\right) \times\left(3 \times 10^{10}\right)}{\left(3 \times 10^{-5}\right) \times\left(1.6 \times 10^{12}\right)}\)eV
= 4.14 eV
Example 4.
Work functions of three metals A, B and C are 1.92 eV, 2.0 eV and 5.0 eV respectively. Which metal will emit photoelectrons when a light of wavelength 4100Å is incident on the metal surfaces? [CBSE ‘o5]
Solution:
Energy of incident photon,
E = hf = h ᐧ \(\frac{c}{\lambda}\) = \(\frac{6.625 \times 10^{-27} \times 3 \times 10^{10}}{4100 \times 10^{-8}}\)erg
= \(\frac{6.625 \times 10^{-27} \times 3 \times 10^{10}}{1.6 \times 10^{-12} \times 4100 \times 10^{-8}}\) eV ≈ 3.03 eV
Hence, this photon will be able to emit photoelectrons from metals A and B but not from C.
Example 5.
In a microwave oven the electromagnetic waves are generated having wavelength of the order of 1 cm. Find the energy of the microwave photon. (h =633 × 10-34J ᐧ s). [WBCHSE Sample Question]
Solution:
A = 1 cm = 10-2m; c = 3 × 108m ᐧ s-1
∴ E = hf = \(\frac{h c}{\lambda}\) = \(\frac{\left(6.63 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{10^{-2}}\)
= 1.989 × 10-23J = \(\frac{1.989 \times 10^{-23}}{1.6 \times 10^{-19}}\)eV = 1.24 × 10-4eV