How many electrons in an atom may have the following quantum numbers? (a) n = 4, ms = -(1)/(2) (b) n = 3, l =

Exercises · 2.31

Q. How many electrons in an atom may have the following quantum numbers?

(a) n = 4,

m_s

-\tfrac{1}{2}

(b) n = 3, l = 0

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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. …

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The key idea is to count electrons by applying the Pauli exclusion principle and the rules for quantum numbers. For (a), all electrons in the n=4 shell with spin down: 16 electrons. For (b), only the 3s subshell: 2 electrons.

Let’s build the intuition first. Quantum numbers are like an address for an electron in an atom. The principal quantum number $n$ defines the shell (energy level). The azimuthal quantum number $l$ defines the subshell (shape). The magnetic quantum number $m_l$ defines the orbital orientation. And the spin quantum number $m_s$ defines the electron’s spin direction — either $+\frac{1}{2}$ or $-\frac{1}{2}$ .

The Pauli exclusion principle says no two electrons in the same atom can have the same set of all four quantum numbers. So each unique combination of $(n, l, m_l, m_s)$ holds at most one electron. Counting electrons means counting how many such combinations satisfy the given conditions.

Now, step by step.

Part (a): n = 4, $m_s = -\frac{1}{2}$ Here, $n$ $n$ is fixed to 4, and spin is fixed to down. But $l$ $l$ and $m_l$ $m_{l}$ can vary — as long as they obey the rules: $l$ $l$ goes from 0 to $n-1$ $n - 1$ , and for each $l$ $l$ , $m_l$ $m_{l}$ goes from $-l$ $- l$ to $+l$ $+ l$ in integer steps. For $n=4$ $n = 4$ , possible $l$ $l$ values: 0, 1, 2, 3.
- For $l=0$ : $m_l$ can only be 0 → 1 orbital.
- For $l=1$ : $m_l = -1, 0, +1$ → 3 orbitals.
- For $l=2$ : $m_l = -2, -1, 0, +1, +2$ → 5 orbitals.
- For $l=3$ : $m_l = -3, -2, -1, 0, +1, +2, +3$ → 7 orbitals. Total orbitals in n=4: $1 + 3 + 5 + 7 = 16$ orbitals. Each orbital can hold exactly one electron with $m_s = -\frac{1}{2}$ (since the other spin is $+\frac{1}{2}$ ). So the number of electrons is exactly the number of orbitals: 16. …

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Q.How many electrons in an atom may have the following quantum numbers?

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