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Planck’s Constant

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Last updated date: 28th Mar 2024
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Introduction

The energy was initially considered to be continuous. However, after a long research Max Planck came to the conclusion that the energy is not continuous in nature it is instead discrete and in small packets which are indicated by the small invisible particles called photons. These particles are the ones that carry the energy and this energy that has been carried is determined by Planck’s Constant. To know more about Planck’s Constant - value, Formula, Symbol, Applications, and Examples students can now check out more about the same via Vedantu.


Planck’s Constant - Value, Formula, Symbol, Applications, and Examples

The energy that is released in the form of packets or chunks in a discontinuous manner is known as Photons where the energy of each photon is directly proportional to the frequency i.e. E and depends upon f.


                     E  ∝  f,   E = k x  h x u….(1)   (k is no of photons, and is an integer)


Here, ‘h’  is called the Planck’s constant.


On this page, we shall learn about the following:


  • Planck’s constant

  • Value of Planck’s constant

  • Deriving Planck’s constant formula in both MKS and CGS unit

  • Planck’s constant units and dimensional formula

  • Planck’s constant symbol

  • Applications of Planck’s constant with examples

  • Illustrative examples to understand the basics of this topic. 


What is Planck's Quantum Theory all about?

A German theoretical physicist, Dr. Max Planck had put forth a theory known as Planck’s quantum theory. This theory states: Energy radiated or enwrapped is not perpetual, but in the form of packets called quanta. This energy is known as the “Quantum of energy.” For a single packet, we call it quanta, where quanta is an integer value, unlike continuous energy supply which has varying values: 1 or 1.1 or 1.2…


Packets are units of energy and they are called Quanta in general terms whereas Photons is a term used for packets in terms of visible light.


Consider this equation:

 E = h x c/λ….(2)


 h = 6.626 x 10⁻³⁴


 c = 3 x 10⁸ m/s


Put this value in the above equation(2)


(6.626 x 10⁻³⁴) * (3 x 10⁸)/λ


(19.878 x 10⁻²⁶)/λ  ∽   (2 x 10²⁵)/λ


We get,


M = (2 x 10²⁵)/λ


This is the value for energy of a single photon, and for ‘k’ no of photons, it would be:


E = (k x 2 x 10²⁵)/λ


The value of E is calculated only when wavelength, λ is given in meters. If λ is given in any other unit let’s say in Angstrom, simply, we can convert 1 Angstrom to meters (1 Angstrom = 10⁻¹⁰m) where h is the Planck’s constant, and h = Energy of a quantum of electromagnetic radiation divided by its frequency.


Planck’s constant ‘h’  is measured in Joule-seconds in the SI system.


       h  = 6.626 x 10⁻³⁴


and Electronvolt or (eV) in the M.K.S system.


1 eV  =  1.6 x 10⁻¹⁹ Joule 


E = (12400/λ) eV for λ in  Å. 

E = (1240/λ) eV for λ in nm.

 

Value for λ when E = 4.13 V 


E = 12400/λ


4.13 = 12400/λ


λ  = 12400/4.13 = 3000 Å


Experiments Used to determine Planck's Constant:

There were two experiments that were used in order to determine the Planck constant and can be provided as follows:


1.Kibble Balance


2.X-ray crystal density method


  1. Kibble Balance:

It is the accurate weighing machine that was named after its inventor Bryan Kibble in 1975. It is designed to equalize one of the forces that arise with another. In this case, the weight of a test mass is exactly balanced out by a force that is induced when an electrical current is run through a coil of wire that has been immersed in a surrounding magnetic field.


  1. X-ray Crystal Density Method:

This method is a primary method that is used in the determination of Planck’s constant. Here crystals of silicon which are available in high quality and purity by the semiconductor industry are used.


What’s so Special about Planck’s Constant?

A blackbody is an idealized physical body, which assimilates all the electromagnetic radiation. Upon heating, it reflects the light falling on it, but that too of varying amounts of wavelengths.


(Image wil lbe Uploaded Soon)


(Image wil lbe Uploaded Soon)


Here in this graph, we can observe that less is the wavelength, lesser is the emission of waves, then a time comes when we get the maximum wavelength, Vmax, which means maximum emission.


The Vmax is the position shown as a peak in the graph is the visible light.


What happens here is when we go further, the wavelength keeps on increasing, but the emission of waves keeps on decreasing and continues further, we see that the emission of waves is negligible, but not zero. (All the wavelengths of whatever amount and irrespective of frequencies are radiated).


But from theoretical derivation, you must have observed in the curve that from the starting till the point when the wavelength is maximum, the graph shows symmetry, but what happens thereafter? The emission of waves is maximum even when the wavelength is less.


There’s a lot of difference when the wavelength is less. The modification in the above concept was brought up by a great German theoretical physicist, named Dr. Max Planck.

Where he considered light as a form of ‘k’ no of chunks or packets called photons by the relation.


E = k x h x f, k being the no of photons.


After his experimentation, the experimental and theoretical curves which were not symmetrical to each other, got into symmetry infers that the theory given by Dr. Planck was correct.


Summary

  • When an iron rod is heated, lights of all wavelengths are emitted, but a human eye can perceive only that light which is of maximum wavelength Vmax. 

  • We estimate the temperature of stars by observing the Vmax of the light emitted.

FAQs on Planck’s Constant

1.What are the applications of Planck’s constant?

Planck’s constant value has been used in a number of applications including Planck’s equation. For example, to find out the energy that is related to a photon the frequency will be multiplied with Planck’s constant to get the desired answer. It is also highly used in a lot of formulae and can be provided as follows:


1. Formula of the spectral radiance of a body in black body radiation can be provided as follows:


\[B_v(v,T)=\frac{2hv^3}{c^2}\frac{1}{\frac{hv}{e^kB^t}-1}\]


2. Relation between Planck Einstein relation can be provided as follows:

E=hf

3. The relation between position and momentum in the uncertainty principle can be provided as follows:


\[\lambda x\lambda p\geq \frac{\frac{h}{2\pi }}{2}\]


4. The de Broglie wavelength through matter-wave equation can be provided as follows:


\[\lambda =\frac{h}{p}\]

2.In the determination of Planck’s Constant why is LED used?

LED is widely used in the determination of Planck’s constant due to the color of LED that it emits when it has a different threshold voltage than the one that produces the electrons. The one that produces the electrons will be of different voltage than the one that emits the light. This voltage when it is combined with the emission wavelength is very useful in determining the Planck's constant. This also allows the Planck’s constant to be determined efficiently without any mistakes occurring as the light that is emitted will only be related to the threshold voltage and hence be accurate.

3.What is a Planck’s mass?

It is seen that there is Planck's Value for all units such as the Planck mass, the Planck charge, the Planck area, the Planck temperature, the Planck angular frequency, the Planck pressure, the Planck impedance, and so on. Out of these, the Planck mass is one of the special cases. This Planck mass is around a milligram and some particles such as the electron neutrino and the electron antineutrino tend to weigh much lower than the Planck mass. However, they tend to exceed the Planck mass when their energy and mass are added together.

4.What is Planck’s law and what is its history?

It is seen that in Physics, Planck's law tells us about the spectral density of electromagnetic radiation that has been emitted by a black body in thermal equilibrium which is at a given temperature T when there has been no net flow of energy or matter between the body and its environment. It was seen that in the late 1800s and the early 1900s, Max Planck and other scientists were trying to understand the difference between the classical mechanics which is the motion of the bodies as explained by Sir Issac Newton in the late 1600s, and a photon which is ultrasmall invisible energy that tends to behave like a wave and a particle.

5.Why is it important to learn about Planck’s Constant - Value, Formula, Symbol, Applications, and Examples through Vedantu?

Planck’s constant is a very important value not only in Physics but in various other subjects including Chemistry and material engineering. This value is used to find out the amount of energy that a photon can carry based on the frequency of the wave in which it travels. This allows us to check the energy and place the values accordingly. This value which is of much importance is hence supposed to be memorized by students for their exams. This can now be possible by checking out the Vedantu NCERT Solutions for Physics for all the students who wish to learn more about the topic Planck’s Constant - Value, Formula, Symbol, Applications, and Examples.

6.Why is Light Called an Electromagnetic Wave?

Light is formed by the combination of the Magnetic and Electric field, and these two fields are perpendicular to each other, which means these two waves are oscillating in a direction perpendicular to each other and the wave is propagating in the direction perpendicular to these two fields.

7.Derive the Dimensional Formula for Planck’s Constant ‘h’.

Dimensional formula for h

Since  h  =  E / f


E  = ML²T⁻²


f= M⁰L⁰T⁻¹  

  

h  = ML²T⁻² / M⁰L⁰T⁻¹ 


ML²T⁻¹

8.Calculate the Ratio of Energies of Two Sources which Emit Light of Wavelength 4000 and 5000 Å to show the Relation of Energy (E) is Inversely Proportional to the Wavelength (λ).

 E1 = (h x v)/λ1  …..(1)


E2 = (h x v)/λ2 ….(2)


eq (1) / eq (2)


We get:  E1/E2 = λ2/λ1 =  4,000/5,000 = 4/5 = 4:5


 We conclude that more is the energy, lesser is the wavelength and vice-versa.

9.A Light Source of λ = 3000 Å Emits 0.6 J of Energy. Calculate the Number of Photons.



Given E = 0.6 J, λ  = 3000 Å ... (1)


We know that  E = (k x h x c)/λ  = (k x 2 x 10⁻²⁵)/v


Now putting the values in eq(1)


0.6 = (k x 2 x 10⁻²⁵)/(3,000 x 10⁻¹⁰)


k = (3,000 x 0.6 x 10⁻¹⁰)/(2 x 10⁻²⁵)


k=  900 x 10¹⁵ = 9 x 10¹⁷ photons

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