I'm a tenth grader who's just been introduced to this subject of atomic structure, so please help.

I recently read that an electron doesn't have circular motion around a nucleus, its motion is kind of bizzare, buzzing and basically all over the place. (As long as it's around the nucleus) But we use the azimuthal quantum number to talk about an electrons orbital angular momentum.

Now, to be sure I didn't know what that meant, so I looked it up. I can't figure out why is there a question of orbital angular momentum if the electron isn't revolving around the nucleus.

Also, is the azimuthal number somehow related to the energy an electron has? I can't figure out if it's the radius that matters or the energy of the electron, so can someone please explain that?

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    $\begingroup$ The picture you have in your mind is the angular momentum of a particle ("revolving around the nucleus"), which is easy to visualise. The problem is that the behaviour of an electron cannot be solely described by treating it as a classical particle. Quantum mechanics is a huge paradigm shift: it's not just an incremental theory that refines the old Bohr model of orbiting electrons and reconciles some problems with it. It's an entirely new way of looking at things, essentially from scratch, and that's why it has the reputation it has now. $\endgroup$ Commented Apr 21, 2018 at 14:43
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    $\begingroup$ @orthocresol that makes sense, thanks. But it's kind of tricky since my course in physics and maths is barely close to explain quantum mechanics. Can you please tell me how should I think of angular momentum of an electron, then? $\endgroup$ Commented Apr 21, 2018 at 16:20
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    $\begingroup$ The electron can have zero or non zero but quantised amounts orbital angular momentum, depending on what state it is in the atom.; s orbital, zero orbital ang. mom., p orbital 1 unit , d two etc. Perhaps the easiest way to deal with this is not to imagine the electron like a ball or fuzzy thing but that it has properties and is described by equations that are indicative of angular momentum. The electron also has spin angular momentum but is not' spinning'. As you realise quantum behaviour is often hard to explain in everyday terms and not just for students :) $\endgroup$
    – porphyrin
    Commented Apr 21, 2018 at 17:28
  • $\begingroup$ Though the concept of angular momentum is used to describe an electron's state, if the electron actually were moving in an orbit, it would radiate energy because it would be an accelerating charge (i.e. centripetally). It's somewhat mind boggling, but an electron can be detected here, or somewhere else, without having traversed the space between the points. Tunnel, or Esaki, diodes provide a different but easily observable example of quantum oddity. en.wikipedia.org/wiki/Tunnel_diode $\endgroup$ Commented Apr 23, 2018 at 0:39

2 Answers 2


In quantum mechanics, the classical concept of trajectory of a particle is abandoned. The quantum state of a particle i.e. the electron is characterized by a wave function $$\psi(\pmb{r}, t)$$ that contains all information about the particle and that extends over space. The simplest case where angular momentum appears in the context of quantum mechanics is the one of a particle in a “ring”. Now you have a wave function which extends over a circular region of space which implies that your particle has angular momentum. When you apply the same idea to the Hydrogen atom you will find the orbital angular momentum.

The energy levels of the Hydrogen atom is given by $E_{n} = E_{0}/n^{2} $, where $E_{0} = 13.6 eV$ and n = 1, 2, 3, ... is the principal quantum number and describes the size of the orbital. Therefore, we see that $E_{n}$ is independent of the orbital angular momentum. If the atom is in the presence of a magnetic field then things change.

  • $\begingroup$ >The simplest case where angular momentum appears in the context of quantum mechanics is the one of a particle in a “ring”. A ring or a sphere? $\endgroup$ Commented Apr 25, 2018 at 7:27
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    $\begingroup$ In a sphere you have two variables in contrast to only one in a ring. The math to work out the particle in a ring is not complicated. Aside from the introductory part of quantization of matter a tenth grade student can understand the problem quite well. $\endgroup$
    – rbw
    Commented Apr 25, 2018 at 12:50
  • $\begingroup$ Ah, misunderstood your expression about the simplest case. You mean the simplest mathematical case and not the ground state of an electron. $\endgroup$ Commented Apr 25, 2018 at 13:11

The concept of orbital angular momentum can be thought of this way.
The electron cloud has a motion surrounding nucleus and its motion can be described by the 'Wave Function' which is nothing but a function of space and time( $\mathrm{\Psi(x,y,z,t))}$. So, the electron cloud has positions described by these wave functions, and also it has definite momentum which is defined as(also, can be derived from $\mathrm{\frac{d<x>}{dt}=<p>}$) $$\mathrm{\hat{p}\Psi = i \hbar (\frac{\partial \Psi}{\partial x}\hat{x} + \frac{\partial \Psi}{\partial y}\hat{y} + \frac{\partial \Psi}{\partial z}\hat{z}) = i \hbar \vec{\nabla}\Psi}$$ So, if you have the wave function of the electron you already have momentum for it. Now, according to the definition of Angular Momentum, we have $L = (\vec{r} \times\vec{p} ) $, and thus we also have Angular momentum of the electrons by this definition where the individual components are, $$\mathrm{\hat{L_x} \Psi= i \hbar(y \frac{\partial \Psi}{\partial z} - z \frac{\partial \Psi}{\partial y})}$$$$\mathrm{\hat{L_y} \Psi= i \hbar(z \frac{\partial \Psi}{\partial x} - x \frac{\partial \Psi}{\partial z})}$$ and also $$\mathrm{\hat{L_z} \Psi= i \hbar(x \frac{\partial \Psi}{\partial y} - y \frac{\partial \Psi}{\partial x})}$$ and similarly, there is also $\hat{L^2} $ operator( square of angular momentum) = $\hat{L_x^2}+ \hat{L_y^2} + \hat{L_z^2}$Now what you know about the Orbital Angular Momentum, is nothing but when we try to find the eigen value of $\hat{L^2} $ by solving the equation $\hat{L^2}\Psi = E \Psi $ From solving that equation, we get some definite eigenvalue of the angular momentum, which is reffered as those Azimuthal Quantum numbers( $l$).Those eigen values are $\hbar^2l(l+1)$, which you might have encountered as the orbital angular momentum of an electron possesing that($l$) particular quantum number. In quantum mechanical term these, are expectation values ( Eigen values are always expectation values) of Orbital Angular momentum.
So, though the electron is not actually circling around the nucleus, it has positions and momentum and that's why Angular momentum according to definition and also, when we are concerned about systems of electrons, the solution corresponds to some finite states of presence of electrons, which are termed as orbitals, and the angular momentum associated to that becomes orbital Angular Momentum.

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    $\begingroup$ I dont know if we can expect a tenth grader to have the math background to fully understand this answer. $\endgroup$
    – KanyeBest
    Commented Apr 23, 2018 at 23:25

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