# Planck’s Quantum Formula

Maxwell’s proposal concerning the wave nature of electromagnetic radiation was useful in describing phenomena such as interference, diffraction, and other phenomena as science progressed. However, he was unable to explain a number of other observations, including the nature of radiation emission from hot bodies, the photoelectric effect, which is the ejection of electrons from a metal compound when electromagnetic radiation strikes it, the dependence of solids’ heat capacity on temperature, and atom line spectra (especially hydrogen), which all were explained by Planck’s Quantum Theory, Let’s learn Planck’s Quantum Theory in Detail.

### Planck’s Quantum Theory

Quantum mechanics natural phenomenon is Planck’s quantum theory. Max Planck, a German physicist, proposed the hypothesis. It clarifies the quantum nature of electromagnetic wave energy. The nature of radiated emission and other phenomena that the laws of classical mechanics could not account for are addressed by Planck’s quantum theory.

Albert Einstein and Max Planck co-created the fundamental physics theories during the 20^{th} century. If Wien’s law had to hold in the high-frequency area, he realized that radiation entropy had to be mathematically dependent on its strength. He also recognized the need of relying on a low-frequency zone for test findings. Planck reasoned that he should combine the two terms in a straightforward manner and convert the result into a formula for radiation in its range. As a result, the ‘Planck Radiation Law’ was founded by him.

Furthermore, Planck had to consider that the oscillators that connect a black body and re-emit the phenomenon of attractive energy over them could not absorb this energy continuously, but rather at different rates, resulting in a quantum of energy; by statistically distributing these quanta, each contains an amount of energy equal to its quantity, on top of all the oscillators available to the black body.

Planck rediscovered the formula he’d discovered two months before. He used it to test the constant h (his result was 6.55 Ã— 10^{-27} erg-second, which is similar to the present value of 6.626 Ã— 10^{-27} erg-second). As a result, he came up with a formula for calculating the radiation’s energy. In today’s world, Planck’s simple assumption of quantization is recognized as the origin of quantum theory. Our understanding of atomic and subatomic processes has been revitalized as a result of this new perspective.

Quantum energy refers to the smallest amount of energy that electromagnetic radiation may release or absorb. In addition, the amount of radiation absorbed or released is proportional to the total amount of radiation.

**Planck’s quantum theory postulates**

Following are postulated of Planck’s quantum theory:

- The energy is not constantly radiated or released. Small amounts of it are released as quanta, which are energy packets.
- Each particle of radiation is known as a photon when it takes on the appearance of light.
- A photon’s or a quantum of energy’s energy is directly proportional to the frequency of the radiation. If Planck’s constant is h, and the frequency of radiation is Î½, together they form the equation
**E = hÎ½.** - The quantum energy is a multiple of a whole number, such as hÎ½, 2hÎ½, 3hÎ½, and so on, which is used to express the total energy of radiation.

### Planck’s Quantum Formula

Radiation’s energy can be calculated as follows:

Each quantum’s energy (E) is proportional to the frequency at which it occurs,

âˆ´ E Î± Î½

E = hÎ½Where,

- E = Radiation Energy,
- h = Planck’s Constant (6.626 Ã— 10
^{-34}Js),- Î½ = Radiation Frequency.

E = hc / Î» â€¦ (Î½ = c/Î»)Where,

- c = Speed of light,
- Î» = Wavelength.
Total energy emitted or absorbed by any particle is an integral multiple of hÎ½.

âˆ´

E = nhÎ½

### Electromagnetic Radiation

Electromagnetic rays are a sort of energy that can be disseminated in a vacuum or via a medium material, and they have been demonstrated to have particle-like structures. Electromagnetic rays include radio waves, microwaves, infrared, visible light, UV, X, and gamma rays. Periodic changes in the electric field or magnetism produce electrical radiation, also known as electric waves. Different wavelengths of the electromagnetic spectrum are created depending on the frequency of these changes and the energy produced. electric waves. All electromagnetic waves travel at the speed of light.

### Black Body Radiation

The threshold frequency is the lowest frequency used for emission. Assume that the light is falling with the energy of hÎ½. The value of ‘hÎ½’ is employed as binding energy in addition to the threshold energy the remainder is provided to the electron as kinetic energy. The following are observed:

- When the frequency exceeds the frequency limit, kinetic energy is used to extract electrons rather than emission.
- There is no discharge and no kinetic energy is given if the frequency is less than the threshold.

The black body is a well-organized object that can absorb any electrical radiation it comes into contact with. Then, depending on its temperature, it starts emitting hot rays in a constant stream. A black body’s radiation is referred to as “black body radiation.” The stars resemble a black body in appearance.

The amount of radiation released by each wavelength’s length is solely determined by the object’s temperature, not by any other factor such as its chemical composition. In 1900, German philosopher Max Planck provided a statistical explanation. Radiation is the term used by scientists to describe this. They also use the term ‘black body’ radiation, which refers to a piece of hypothetical equipment that absorbs all light and emits nothing. Overall wavelengths, the black body is a perfect light bulb, however, there is one wavelength where radiation is very powerful.

### Relation of black body radiation with Planckâ€™s law

Temperature and radiation are related, according to Planck’s radiation law. The radiation that a black body emits at all wavelengths increases together with its temperature. The relationship is provided by:

B(Î½, T) = (2hÎ½^{3}/c^{2}) Ã— (1/(e^{hÎ½/k}_{b}^{T}– 1))Where,

- Î½ = Frequency,
- k
_{b}= Boltzmann Constant,- h = Planck’s Constant,
- c = speed of light in vacuum.

### Applications of Planck’s Quantum formula

Quantum mechanics is based on Planck’s quantum theory, which is the most fundamental theory. As a result,

- It can be applied in any sector that employs quantum physics.
- It finds uses in electrical appliances, medicine, quantum computers, lasers, and quantum cryptography, among other things.

### Sample Problems

**Problems 1: Two electrons have wavelengths of 1352 angstroms and 4056 angstroms, respectively. Determine the energy ratio between them.**

**Answer**:

Since,

E = hc/Î»

Therefore, E is inversely proportional to Î». As a result, the ratio of energy to wavelength will be the inverse.

âˆ´ E

_{1}/E_{2}= Î»_{2}/Î»_{1}âˆ´ E

_{1}/E_{2}= 4056/1352

âˆ´ E_{1}/E_{2}= 3

**Problems 2: Calculate the energy associated with a 55 nm wavelength.**

**Answer**:

Since,

E = hc/Î»

âˆ´ E = 6.63 Ã— 10

^{-34}Ã— 3 Ã— 10^{8}/ 55 Ã— 10^{-9}âˆ´ E = 19.89 Ã— 10

^{-26}/ 55 Ã— 10^{-9}

âˆ´ E = 36.16 Ã— 10^{-19}J_{ }_{ }

**Problems 3: Calculate energy radiation with a radiation frequency of 34 Hz.**

**Answer**:

Since,

E = hÎ½

âˆ´ E = 6.63 Ã— 10

^{-34}Ã— 34

âˆ´ E = 225.42 Ã— 10^{-34}J

**Problems 4: Calculate wavelength when energy radiation is 21.9 Ã— 10 ^{-15} J.**

**Answer**:

Since,

E = hc/Î»

âˆ´ Î» = hc/E

âˆ´ Î» = 6.63 Ã— 10

^{-34}Ã— 3 Ã— 10^{8}/ 21.9 Ã— 10^{-15}âˆ´ Î» = 19.89 Ã— 10

^{-26}/ 21.9 Ã— 10^{-15}

âˆ´ Î» = = 0.9082 Ã— 10^{-11}m

**Problems 5: The wavelengths of two electrons are 421 angstroms and 729 angstroms, respectively. Calculate the energy balance between them.**

**Answer**:

Since,

E = hc/Î»

Therefore, E is inversely proportional to Î». As a result, the ratio of energy to wavelength will be the inverse.

âˆ´ E

_{1}/E_{2}= Î»_{2}/Î»_{1}âˆ´ E

_{1}/E_{2}= 729/421

âˆ´ E_{1}/E_{2 }= 1.73

**Problems 6: If radiation frequency is 20 Hz then calculate the radiation energy.**

**Answer:**

Since,

E = hÎ½

âˆ´ E = 6.63 Ã— 10

^{-34}Ã— 20âˆ´

E = 132.6 Ã— 10^{-34}J

**Problems 7: When the energy radiation is 55.1 Ã— 10 ^{-9} J, calculate the wavelength.**

**Answer**:

Since,

E = hc/Î»

âˆ´ Î» = hc/E

âˆ´ Î» = 6.63 Ã— 10

^{-34}Ã— 3 Ã— 10^{8}/ 55.1 Ã— 10^{-9}âˆ´ Î» = 19.89 Ã— 10

^{-26}/ 55.1 Ã— 10^{-9}âˆ´

Î» = = 0.3609 Ã— 10^{-17}m

**Problems 8: Determine the energy corresponding to a wavelength of 21 nm.**

**Answer:**

Since,

E = hc/Î»

âˆ´ E = 6.63 Ã— 10

^{-34}Ã— 3 Ã— 10^{8}/ 21 Ã— 10^{-9}âˆ´ E = 19.89 Ã— 10

^{-26}/ 21 Ã— 10^{-9}

âˆ´ E = 0.9471 Ã— 10^{-17}J

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