# give de broglie’s explanation of bohr’s angular momentum quantisation postulate.

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## Angular Momentum Of Electron

The angular momentum of an electron by Bohr is given by mvr or nh/2π. According to Bohr’s atomic model, the angular momentum of electrons orbiting around the nucleus is quantized. Louis De Broglie further explained the quantisation of angular momentum.

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## Angular Momentum Of Electron

In this article, we will be learning about the angular momentum of electrons in detail. We will also be learning about De Broglie’s Explanation of the Quantization of Angular Momentum of Electrons.

**Table of Contents**

What is the Angular Momentum of Electron?

De Broglie’s Explanation to the Quantization of Angular Momentum of Electron

Frequently Asked Questions – FAQs

## What is Angular Momentum of Electron?

Bohr’s atomic model laid down various postulates for the arrangement of electrons in different orbits around the nucleus. According to Bohr’s atomic model, the angular momentum of electrons orbiting around the nucleus is quantized. He further added that electrons move only in those orbits where the angular momentum of an electron is an integral multiple of h/2. This postulate regarding the quantisation of angular momentum of an electron was later explained by Louis de Broglie. According to him, a moving electron in its circular orbit behaves like a particle-wave.

The angular momentum of an electron by Bohr is given by **mvr** or **nh/2π **(where v is the velocity, n is the orbit in which the electron is revolving, m is mass of the electron, and r is the radius of the nth orbit).

**Among the various proposed models over the years, the Quantum Mechanical Model seems to best fit all properties. Watch the video to learn more about the features of the quantum mechanical model.**

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### De Broglie’s Explanation to the Quantization of Anfgular Momentum of Electron:

The behaviour of particle waves can be viewed analogously to the waves travelling on a string. Particle waves can lead to standing waves held under resonant conditions. When a stationary string is plucked, a number of wavelengths are excited. On the other hand, we know that only those wavelengths survive which form a standing wave in the string, that is, which have nodes at the ends.

Thus, in a string, standing waves are formed only when the total distance travelled by a wave is an integral number of wavelengths. Hence, for any electron moving in kth circular orbit of radius rk, the total distance is equal to the circumference of the orbit, 2πrk.

**2πrk**= kλ

Let this be equation (1).

Where,

**λ**is the de Broglie wavelength.

We know that de Broglie wavelength is given by:

**λ =**h/p

Where,

**p**is electron’s momentum

**h**= Planck’s constant

Hence,

**λ**= h/mvk

Let this be equation (2).

Where** mvk** is the momentum of an electron revolving in the kth orbit. Inserting the value of λ from equation (2) in equation (1) we get,

**2πrk**= kh/mvk

**mvkrk**= kh/2π

Hence, de Broglie hypothesis successfully proves Bohr’s second postulate stating the quantization of angular momentum of the orbiting electron. We can also conclude that the quantized electron orbits and energy states are due to the wave nature of the electron.

**Related Articles**

What is Angular Momentum?

Bohr’s Atomic Model Energy Level

## Frequently Asked Questions – FAQs

### Is it possible for electrons to have angular momentum?

Yes, it is possible for electrons to have angular momentum.

### How to find the angular momentum of an electron?

The angular momentum of an electron by Bohr is given by mvr or nh/2π (where v is the velocity, n is the orbit in which electron is revolving, m is mass of the electron, and r is the radius of the nth orbit).

### Is the Angular momentum of an electron quantised?

According to Bohr’s atomic model, the angular momentum of electrons orbiting around the nucleus is quantized. He further added that electrons move only in those orbits where the angular momentum of an electron is an integral multiple of h/2.

### What is Bohr’s atomic model?

In the year 1913, Niels Bohr proposed an atomic structure model, describing an atom as a small, positively charged nucleus surrounded by electrons that travel in circular orbits around the positively charged nucleus with attraction provided by electrostatic forces, popularly known as Bohr’s atomic model.

### Who proposed the quantization of angular momentum?

Bohr proposed the quantization of angular momentum.

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## De Broglie's Explanation of Bohr's Second Postulate

Read formulas, definitions, laws from Bohr's Model here. Click here to learn the concepts of De Broglie's Explanation of Bohr's Second Postulate from Physics

Class 12 >>Physics >>Atoms

>>Line Spectra of the Hydrogen Atom

>>De Broglie's Explanation of...

## De Broglie's Explanation of Bohr's Second Postulate

Physics 44608 Views

## DEFINITION

### De Broglie's Justification of Bohr's Assumption

De Broglie came up with an explanation for why the angular momentum might be quantized in the manner Bohr assumed it was. De Broglie realized that if you use the wavelength associated with the electron, and assume that an integral number of wavelengths must fit in the circumference of an orbit, you get the same quantized angular momenta that Bohr did.

The circumference of the circular orbit must be an integral number of wavelengths:

2πr=nλ= p nh (λ= p h )

The momentum, p, is simply mv as long as we're talking about non-relativistic speeds, so this becomes:

2πr= mv nh

Rearranging this a little gives the Bohr relationship:

L r =mvr= 2π nh

### REVISE WITH CONCEPTS

## Line Spectra of the Hydrogen Atom

Example Definitions Formulaes > IMPORTANT QUESTIONS

### Calculate the shortest wavelength in the Balmer series of hydrogen atom. In which region (infra-red, visible, ultraviolet) of hydrogen spectrum does this wavelength lie?

The shortest wavelength present in the Paschen series of spectral lines in hydrogen atom is nearly (Rydberg's constant, R=1.09710Easy View solution > 7 1/m) : Medium View solution >

### The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is

Medium NEET View solution >

### If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength λ. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be

The ratio of de-Broglie wavelength of molecules of hydrogen and helium in two gas jars kept separately at temperature 27Show that the radius of the orbit in hydrogen atom varies as nMedium View solution > o C and 127 o C respectively is Medium View solution > 2

. Where n is the principal quantum number of the atom

Medium View solution >

### The shortest wavelength of the Brackett series of a hydrogen-like atom (atomic number Z) is the same as the shortest wavelength of the Balmer series of hydrogen atom. Find the value of Z.

If an electron jumps from the 4th orbit to the 2nd orbit of hydrogen atom, then the frequency of emitted radiation in the hertz will be:Medium View solution >

(Take Rydberg's constant, R=10

The de-Broglie wavelength of an electron in 45 cm −1 ) Medium View solution > th

orbit is ________. (r= radius of 1

st orbit) Easy View solution >

### Maximum wavelength in balmer series of hydrogen spectrum is

Medium View solution >

## Given de

For an electron moving in n^(th) drcular orbit of radius r(n), the total distance is =2pir(n) Circumference of a stationary Bohr orbit of radius r(n) is equal to integral multiple of wavelength of matter waves 2pir(n)=nlamda" "......(i) The de-broglie wavelength of the electron moving in the n^(th) orbit. lamda=(h)/(mv)" "......(2) From equation (1) and (2), 2pir(n)=(nh)/(mv) i.e., mvr(n)=(nh)/(2pi) But, angular momentum of the electron is L=mvr(n) Hence L=(nh)/(2pi)

Given de- Broglie's explanation of quantisation of angular momentum as proposed by Bohr.

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Updated On: 27-06-2022

Text Solution Solution

For an electron moving in

n t h

drcular orbit of radius

r n

, the total distance is

= 2 π r n

Circumference of a stationary Bohr orbit of radius

r n

is equal to integral multiple of wavelength of matter waves

2 π r n = n λ ... ... ( i )

The de-broglie wavelength of the electron moving in the

n t h orbit. λ = h m v ... ... ( 2 )

From equation (1) and (2),

2 π r n = n h m v i.e., m v r n = n h 2 π

But, angular momentum of the electron is

L = m v r n Hence L = n h 2 π Answer

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