Planck’s Quantum Theory: Quantization of Energy

In this chapter, you will learn about the Planck’s quantum theory. Before going to learn about this theory, we will understand some few things.

The wave nature of electromagnetic radiation could successfully explain some of the experimental phenomenon such as diffraction, interference and polarization, based on classical Maxwell’s wave theory of radiation. However, classical wave theory could not explain certain experimental observations, such as:

• the nature of emission of radiation from hot bodies (blackbody radiation).
• the ejection of electrons from the metal surface when light (radiation) strikes it (photoelectric effect).
• variation of heat capacity of solids as a function of temperature.
• the discrete nature of line spectra of atoms with special reference to hydrogen.

The first failure of the wave theory of light was found in the study of black body radiation in 1900. However, the first clear-cut case of failure of wave theory of light was found in the photoelectric effect when Heinrich Hertz observed in 1887. This theory could not explain why the emission of electrons from a metal surface depended on the frequency of light rather than its intensity.

Black-body Radiation

An ideal object that perfectly absorbs all incident radiation, regardless of frequency, and emits radiation at all frequencies when heated is called black-body. When a block-body is heated, it emits electromagnetic radiations, which are called black-body radiations. The colour of these radiations changes from red to yellow to white when the temperature of the black-body increases. At a high temperature, a portion of emitted radiations lie in the visible region of the spectrum.

When you further increase the temperature of the body, the proportion of higher frequency (shorter wavelength) radiation increases. Therefore, when you heat iron, its color changes from reddish to bright red, then to white, and eventually to blue with further heating. The exact frequency distribution of the emission of blackbody radiations from a hot black-body depends upon the temperature of the blackbody.

The emission of radiation from the black-body could not be fully explained based on classical wave theory of Maxwell.

Max Planck’s Quantum Theory of Radiation

In order to explain the phenomenon of black-body radiation and photoelectric effect, a German physicist Max Planck in 1901 put forward a new theory which is known as Planck’s Quantum Theory of Radiation. The postulates of this theory are as follows:

(1) A radiation is associated with a certain amount of energy, which is called radiation energy. Since light and heat are forms of radiation, they are also associated with energy.

(2) Radiant energy is not emitted or absorbed by a body continuously (i.e. not as waves as suggested by the classical Maxwell’s wave theory) but discontinuously (i.e. discretely) in the form of small packets (or bundles) of energy called quantum (in plural quanta). In the case of light, the quantum of energy is called a photon. Thus, a quantum can be defined as:

A quantum is the smallest unit of radiation energy (i.e. radiant energy) which can exist independently.

The below figure shows the emission of radiant energy in a continuous and discontinuous fashion from a heated iron ball.

In the case of light, which is also a form of radiation, light energy is emitted or absorbed in the form of packets or bundles. Each packet or bundle of energy is called a photon (instead of quantum). A photon is a massless bundle of energy. The energy associated with the light radiation is called light energy.

Thus, according to Planck’s quantum theory of radiation, light is composed of massless particles called photons. Under the normal conditions, the presence of photons in the light cannot be detected readily. As far, even light of very low intensity contains billions of photons.

In 1905, Einstein, when he was explaining the concepts of photoelectric effect, had proposed the existence of photons in the light. This was strong evidence for the existence of photons and the particle nature of light. The explanation of Einstein built on Planck’s quantum theory and earned him the Nobel Prize in physics in 1921.

(3) The energy associated with each quantum or photon (E) of a given light or radiation is directly proportional to the frequency of the emitted radiation.

Thus, E ∝ ν

or, E = hν

Here, h is a proportionality constant which is called Planck’s constant. Its numeric value is 6.624 * 10^-27 ergs sec or 6.624 * 10^-34 joule sec.

This equation is Planck’s equation which is applicable to all types of radiations. This equation proves that the energy associated with a photon, E is equal to hν.

Since v = c/λ, where c is velocity of light and λ is wavelength of the radiation.

Thus, E_photon = hc/λ

This formula shows that the energy associated with a quantum or photon of radiation is inversely proportional to its wavelength. Thus, the smaller the wavelength (or higher the frequency) of radiation, larger the energy associated with a photon or quantum.

(4) The energy emitted or absorbed by a body can be either one quantum or any whole number multiple of quanta (hν).

Thus, Energy emitted or absorbed (E) = nhν

Where n is a whole number 1, 2, 3, 4, . . . . etc. but never fractional values like 1.5, 2.5, etc. Thus, the energy emitted or absorbed is quantized.

Albert Einstein in 1905 pointed out that light can be considered to consists of a stream of particles called photons. These photons can be treated as particles of light. According to Einstein, the energy of photon is also related to mass according to famous relation E = mc², where m is the mass of photon. In this way, it was proved that light has wave as well as particle characteristics. That is light has dual character or nature. This duality is a fundamental concept in quantum mechanics.

Solved Examples based on Planck’s Quantum Theory

Example 1: What is the ratio of their energies if the wavelengths of two photons are 2000 Å and 4000 Å respectively?

Solution:

E₁ = hc/λ₁ and E₂ = hc/λ₂

E₁ / E₂ = (hc/λ₁) * (λ₂/hc) = λ₂/λ₁ = 4000 / 2000 = 2 (Ans.)

Example 2: Calculate the energy of a photon of light with a frequency of 1.0 * 10¹⁵ sec^-1.

Solution:

The energy of a photon is given by

E = hν = 6.63 * 10^-34 Joule s * 1.0 * 10¹⁵ s^-1 = 6.63 * 10^-19 J (Ans.)

Example 3: If a light radiation consists of 3×10¹⁸ photons and produces 1.5 J of energy, what is the wavelength of the light radiation?

Solution:

The energy of a photon is given as E = hν.

Energy of N photons E = nhν = nhc/λ

λ = nhc/E = (3 * 10¹⁸ * 6.63 * 10^-34 * 3 * 10⁸) / 1.5 = 39.78 * 10^-8 m = 39.78 * 10^-8 * 10¹⁰ Å

Wavelength of the light radiation λ = 3978 Å (Ans.)

Example 4: Calculate the number of photons emitted by the bulb in one minute if 100 watt bulb is emitting monochromatic light having wavelength of 6000 Å.

Solution:

100 watt means the bulb is emitting 100 Joules per second.

Energy emitted in one minute = 100 * 60 J = 6000 J

Energy of a photon is given by the relation, E = hν = hc/λ

E = (6.62 * 10^-34 * 3 *10⁸) / 6000 * 10^-10 = 3.31 * 10^-19 Joules.

Number of photons emitted per minute = Energy emitted per minute / Energy of a photon

Number of photons emitted per minute = 6000 J / 3.31 * 10^-19 Joules = 1.81 * 10²² photons (Ans.)

Example 5: What is the energy of one mole of photons of radiation having a frequency of 5 * 10¹⁴ Hz?

Solution:

Energy of one photon E = hν = 6.62 * 10^-34 * 5 * 10¹⁴ = 33.1 * 10^-20 Joules

Energy of one mole of photons, E = 33.1 * 10^-20 * 6.022 * 10²³ = 199.51 KJ mol^-1.