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A scientist known as Bohr, has put forward the hypothesis of angular momentum of an electron. If we follow Bohr’s atomic model, we can study the different arrangements of electrons that are located among different orbits near the nucleus.

Also, using Bohr’s atomic model, we can manipulate the angular momentum of the electron that is quantized, revolving around the nucleus.

In his further statement, he included the movement of electrons by stating that the only electron which can revolve around its orbit possesses the angular momentum of its integral multiple of h/2.

Later on, Louis de Broglie explained the postulate of the quantization of the angular momentum of an electron given by Bohr. He stated that an electron in motion shows the behavior of a particle-wave.

Mathematically,

Angular momentum = mvr or nh/2π

Here,

v = velocity

n = the orbit where electrons are present

m = mass of the electron

r = the radius of the nth orbit

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Louis de Broglie explained the quantization of angular momentum of the electron. According to him, the particle waves travel in a cord and have an analogous nature (behavior).

Under the resonant conditions, particle waves can drive towards the standing waves. If we pull out a stationary string plucked, the excitement of wavelength occurs. As per the findings, it is inferred that the waves that have nodes can survive the angular momentum, and later on, they create standing waves in the string.

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Angular Momentum of Electron Formula

Standing waves are formed in a string when a wave completes the total distance of its integral number of wavelengths.

That is why, for an electron in motion in the kth circular orbit, having the radius of rk, the total distance covered by any electron = circumference of the orbit of an electron.

The formula of orbital angular momentum is given by:

2πrk = kλ …. [1]

Here,

λ = de Broglie wavelength

According to the de Broglie equation, the expression is:

λ = h/p

Here,

h is Planck’s constant

p is the momentum of an electron

Therefore, λ = h/mvk …. [2]

Here,

mvk is momentum of electron in the kth orbit

So, by putting the value of λ from equation [2] in equation [1], we get:

kh/mvk = 2πrk

kh/2π = mvkrk

So, we can infer that Bohr’s second postulate is successfully proven by de Broglie hypothesis.

The above inference states the quantization of the angular momentum of the orbiting electron. We can also find the nature of the wave of an electron, which explains the energy states of the electron along with the quantized electron orbits.

The orbital angular momentum of an electron can be obtained using the right formula. To make the learning easier; the orbital angular momentum is quantized. We need to apply the orbital angular momentum quantum number l.

Here is the formula for orbital angular momentum:

L = \[\sqrt{l(l+1)h}\]

We can calculate the angular momentum of a particle having the mass M with radius and velocity v.

The expression is given as:

L = mvr sin θ

Here,

θ = tangential angle at a certain point with the circumference of the orbit.

If we do the vector product, then L = r\[^{\rightarrow }\] х p\[^{\rightarrow }\].

The angular momentum magnitude of a ring-shaped orbit, L = Mvr

We can calculate the orbital angular momentum by using the following orbitals taken as an example:

Example: 1

Find out the orbital angular momentum of these orbitals: 3p, 3d, and 3s.

Ans: The formula to calculate the orbital angular momentum is:

L = \[\sqrt{l(l+1)h}\]

Here, h = \[\frac{h}{2\pi }\] , l = angular momentum

The value of l for different orbitals are given below.

s-orbital = 0

p-orbital = 1

d-orbital = 2

f-orbital = 3

Here, For 3p, so Angular momentum = \[\sqrt{l(l+1)h}\] = \[\sqrt{2h}\]

For orbital 3d, angular momentum = \[\sqrt{2(2+1)h}\] = \[\sqrt{6h}\]

For orbital 3s, angular momentum = \[\sqrt{0h}\] = 0

We can calculate the angular momentum by applying de Broglie’s hypothesis upon an electron of a Hydrogen atom.

We know that the associated wavelength of any particle according to Broglie hypothesis is:

λ = \[\frac{h}{M_{v}}\].....(1)

From classical mechanics, the orbital angular momentum is:

L = Mvr …. (2)

It is also noted that

2πr = mλ

λ = \[\frac{2\pi r}{m}\]

By inserting the value of in equation 1, we get:

\[\frac{2\pi r}{m}\] = \[\frac{h}{mv}\]....[3]

From equation 1 we can get:

M\[_{v}\] = \[\frac{L}{r}\]

By substituting the Mv in equation 3 we get:

\[\frac{2\pi r}{m}\] = \[\frac{h}{(\frac{L}{r})}\] = \[\frac{rh}{L}\]

Now, solve this relation for angular momentum L:

L = rh х \[\frac{m}{2\pi r}\]

or, L = \[\frac{mh}{2\pi }\]

or, L = m . \[\frac{h}{2\pi }\]

Here, \[\frac{h}{2\pi }\] is designated as h.

So, the angular momentum of the electron orbiting in the Hydrogen atom is:

L = m.h

Here, m = 1, 2, 3, 4 …

Also, the angular momentum is quantized in units of ħ.

FAQ (Frequently Asked Questions)

Q1. Is it Possible for Electrons to Have Angular Momentum?

Ans: Yes, the electron does possess angular momentum. The angular momentum of an electron can be found as per the relation given by Bohr.

L = mvr or nh/2π

Here, n = the orbit at which the electron is present

v = velocity

m = mass of the electron

Q2. Give Your Opinion about this Statement “Angular Momentum is a Vector Quantity.”

Ans: Yes, the angular momentum is a vector quantity. Also, angular velocity is a vector quantity. The reason is that they both possess magnitude as well as direction. Here, the direction of the angular momentum acts perpendicularly to the plane of rotation.

Q3. Is the Angular Momentum a Conserved Quantity?

Ans: Yes, the angular momentum is a conserved quantity, as it is a major quantity. According to the physicists, the angular momentum is the rotational equivalent of the linear momentum. Therefore, in a closed-loop/structure, the total angular momentum is said to be constant.

Q4. Explain the Relation Between Angular and Linear Momentum.

Ans: Linear momentum is the product of mass and velocity, whereas the angular momentum is the product of mass and angular velocity. The unit is measured in kilogram meter square.