Orbital angular momentum depends on _____. (i) l (ii) n and l (iii) n and m (iv) m and s

NCERT Exemplar · Q12

Q. Orbital angular momentum depends on __________.

(i) l

(ii) n and l

(iii) n and m

(iv) m and s

Tripura TbseMCQ· 1mImportance★★★★★est

69% · 97/140 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Energy Level Quantization

Energy Level Quantization: From Intuition to Precision

Imagine you're climbing a smooth ramp. You can stop at any height — 1 metre, 1.5 metres, 2.1 metres — anywhere you like. That's how we intuitively think about energy in everyday life: continuous, like a slide.

Now imagine a staircase. You can stand on step 1, step 2, or step 3 — but you cannot stand halfway between step 2 and step 3. There's no such place. The steps are discrete, not continuous.

Energy level quantization is the idea that in the microscopic world of atoms and molecules, energy behaves like a staircase, not a ramp. Electrons in an atom cannot have just any energy — they can only occupy specific, allowed energy levels. Everything else is forbidden.

Why does this happen? The core intuition

In the classical world, an electron orbiting a nucleus would continuously radiate energy, spiral inward, and crash — atoms would be unstable. But atoms are stable. Nature solved this problem by imposing a rule: the electron's angular momentum (and therefore its energy) can only take certain discrete values.

Think of a guitar string. It can only vibrate at specific frequencies — its fundamental and harmonics. You can't pluck it to produce a frequency halfway between two harmonics. The string's vibration is quantized by its boundaries. Similarly, an electron bound to a nucleus is confined in space, and that confinement forces its energy to be quantized.

Note

Quantization is not a mysterious extra rule — it emerges naturally whenever a wave (like an electron's matter wave) is confined. Confinement creates standing waves, and standing waves only exist at specific frequencies.

The precise statement

For a bound system (like an electron in an atom), the total energy $E$ of the system can only take certain discrete values:

E = E_1, E_2, E_3, \dots

where each $E_n$ is a specific, fixed number. The integer $n$ (1, 2, 3, …) is called the principal quantum number. The lowest energy level ( $n=1$ ) is the ground state; higher levels ( $n>1$ ) are excited states.

For the hydrogen atom, the allowed energies are given by:

E_n = -\frac{13.6 \text{ eV}}{n^2}

So:

$n=1$ : $E_1 = -13.6$ eV (ground state)
$n=2$ : $E_2 = -3.4$ eV
$n=3$ : $E_3 = -1.51$ eV
and so on, approaching 0 eV as $n \to \infty$ (the ionization limit)

Important

The negative sign means the electron is bound to the nucleus. Zero energy corresponds to the electron being free (ionized). The more negative the energy, the more tightly bound the electron.

How do we know this is real?

The most direct evidence comes from atomic spectra. When an electron jumps from a higher energy level to a lower one, it emits a photon of light with energy exactly equal to the difference:

\Delta E = E_{\text{higher}} - E_{\text{lower}} = hf

where $h$ is Planck's constant and $f$ is the frequency of the emitted light. …

Start your 14-day free trial to unlock the full solution →

Orbital angular momentum is a purely spatial quantum property, determined only by the azimuthal quantum number $l$ . The correct answer is (A).

The question asks what the orbital angular momentum of an electron depends on. This is a fundamental idea in quantum mechanics and atomic structure — and it's one that students often confuse with the total energy or the magnetic moment.

Let’s start with the concept. In the quantum model of the atom, an electron’s motion around the nucleus is described by a set of quantum numbers. The orbital angular momentum is the angular momentum associated with the electron’s orbital motion (its path around the nucleus, in the classical picture). But in quantum mechanics, this is quantized — it can only take certain discrete values.

The magnitude of the orbital angular momentum is given by:

L = \sqrt{l(l+1)} \, \hbar

where $l$ is the azimuthal quantum number (also called the orbital quantum number), and $\hbar = h/2\pi$ is the reduced Planck constant. Notice that this expression contains only $l$ — no $n$ , no $m$ , no $s$ .

Now let’s walk through the reasoning step by step.

The quantum numbers and their roles
- $n$ (principal quantum number) determines the energy level and roughly the size of the orbital.
- $l$ (azimuthal quantum number) determines the shape of the orbital and the orbital angular momentum.
- $m$ (magnetic quantum number) determines the orientation of the orbital in space, i.e., the $z$ -component of angular momentum.
- $s$ (spin quantum number) describes the electron’s intrinsic spin, not its orbital motion.
Why $n$ doesn’t appear

The energy of an electron in a hydrogen-like atom depends on $n$ (and $l$ for multi-electron atoms), but the orbital angular momentum is a separate property. For a given $l$ , the magnitude $L$ is fixed regardless of which energy level ( $n$ ) the electron is in. For example, a $2p$ electron ( $n=2, l=1$ ) and a $3p$ electron ( $n=3, l=1$ ) both have the same orbital angular momentum: $\sqrt{2}\,\hbar$ .
Why $m$ doesn’t appear …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ11 — Total number of orbitals associated with third shell will be _____. (i…→Next questionQ13 — Chlorine exists in two isotopic forms, Cl-37 and Cl-35 but its atomic …

Q.Orbital angular momentum depends on __________.

🎓Unlock everything free for 14 days

Unlock everything free for 14 days