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Momentum explains the multiplication of velocity and mass can give rise to the momentum of a body. When a body is in motion with a certain amount of mass, it possesses momentum. Well, angular momentum and momentum both are different by their different functionality.

Electrons are always revolving around the nucleus. It possesses angular momentum. So, the angular momentum of an electron is the rotation or the spinning of the electron around the nucleus.

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The angular momentum of the electron in d orbital is equal to âˆš6 (h/2Ï€). We obtain the result with the help of the formula âˆšl (l+1) h/2Ï€. Here, l = 2 as it is d-orbital.Â

It was Bohr who put forward the formula for the calculation of the angular momentum of an electron.

According to Bohr, the formula is mvr or nh / 2Ï€Â

Here,Â

v = the velocity

n = the orbit in which electron is present

m = mass of the electronÂ

r = the radius of the nth orbit

Bohrâ€™s atomic model is responsible for the generation of different postulates that are based on the arrangement of electrons.

Those arrangements of electrons can vary as they are in different orbits around the nucleus.Â

Bohrâ€™s atomic model has some specific comments on the angular momentum of electrons. Bohr suggested that the electrons that are revolving around the nucleus are quantized.Â

He had some additional concepts over the orbiting of electrons around the nucleus. He made a statement that electrons can travel to the selected orbits only where an electronâ€™s angular momentum is an integral product of h/2.Â

Bohr's statement was not good enough to postulate the reason for the angular momentum and orbiting of electrons around the nucleus.

Later, the quantization of the angular momentum of an electron was perfectly analyzed and put forward by Louis de Broglie. He suggested that an electron in motion within its circular orbit acts like a particle-wave.

Example: Find Out the Answer for the Statement â€“ â€˜Angular Quantum Number for d Electron isâ€™

Ans: The orbital angular momentum is (L) = \[\sqrt{l(l+1)}\] \[\frac{h}{2 \pi}\]Â

Here, the orbital d is used. So, the value of l = 2

Therefore, the orbital angular momentum can be represented asÂ

\[\sqrt{6}\] \[\frac{h}{2 \pi}\]Â

De Broglie had mentioned a statement that was based on the quantization of the Angular Momentum of Electron. The particle waves have certain behaviour that can be observed analogously. This procedure helped him understand the motion of the waves that are roaming on a string.Â

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Particle waves are the types of waves that lead towards the standing waves. These waves are apprehended under different resonant conditions. When you try to pluck a stationary string, then a certain amount of wavelengths is getting excited.Â

When we consider the fact differently, we will recognize that certain wavelengths only survive to give rise to angular momentum.Â

All of these standing waves have nodes at the ends.

When standing waves are prepared, the total distance covered by a wave is an integral number of wavelengths.Â

So, if we calculate the total distance covered by the electron when it is moving in the kth circular orbit having the radius rk can be written as:

2Ï€rkÂ = kÎ» â€¦. [1]

Here, Î» = de Broglie wavelength

As per de Broglie wavelength, the equation is

Î» =Â h/p

Here, h = Planckâ€™s constant

p = electronâ€™s momentum

So, Î»Â = h/mvk â€¦. [2]

Here, mvkÂ = the momentum of an electron orbiting around the kthÂ orbit.

So, if we put together two equations 1 and 2, we will find:

2Ï€rkÂ = kh / mvk

mvkrkÂ = kh / 2Ï€

In the above process, de Broglieâ€™s hypothesis has proven the second postulate of Bohr successfully. The hypothesis from de Broglie has made its start from the quantization of angular momentum of the revolving electron. Students can gain another conclusion about the quantized electron orbits and energy states. Both of them are present because of the wave nature of the electron.

The answer to the question is very simple. Here is the solution:

We know that orbital angular momentum can be = \[\sqrt{l(l+1)}\] \[\frac{h}{2 \pi}\]Â

If the orbit is p, then l = 1

So, the orbital angular momentum of P electron = \[\sqrt{l(l+1)}\] \[\frac{h}{2 \pi}\] = \[\frac{h}{2 \pi}\]Â

Therefore, \[\frac{h}{2 \pi}\] is the answer that stands correct for the calculation of orbital angular momentum of P electron.

Electron vortex beams are mostly used in different industries. Their applications include mapping of magnetization, identification of crystal chirality and studying chiral molecules and chiral Plasmon resonances.

We use two methods for measuring orbital angular momentum; they are:

Wavefront flattening

Cylindrically symmetric Stern-Gerlach-like measurement

FAQ (Frequently Asked Questions)

1. Calculate the Angular Momentum of an Electron for an S Orbit.

Ans: As we know, the angular momentum of an orbital (L)= âˆšl(l + 1) (h/2Ï€)Â

For s â€“ orbit, l = 0.

So, L = 0

We conclude that the orbital angular momentum = 0

2. Calculate the Angular Momentum of the Electron in the 5^{th} Orbit as Per Bohrâ€™s Theory.

Ans: As we know, the angular momentum of an orbital (L) = nh/2Ï€Â

The above expression is an integral multiple of h/2Ï€.

So, the orbit is 5. Then, n = 5

We conclude that the orbital angular momentum of the electron in the fifth orbit isÂ L = 5h/2Ï€ = 2.5 h/Ï€Â

3. How Does the Angular Momentum Act in a Conserved System?

Ans: Angular momentum inside the closed system is conserved in all directions after the collision. In this scenario, momentum is conserved. Angular momentum is also the part of the momentum in a collision.Â

This is generated due to the collision of momentum-based electrons with the wall. The angular momentum promotes the spin of electrons after the collision.

4. What Option Do You Choose for the Angular Momentum When Said it is a Scalar or a Vector Quantity?

Ans: You should choose the vector quantity as the angular velocity has the magnitude and direction.