Total number of orbitals associated with third shell will be _____. (i) 2 (ii) 4 (iii) 9 (iv) 3

NCERT Exemplar · Q11

Q. Total number of orbitals associated with third shell will be __________.

(i) 2

(ii) 4

(iii) 9

(iv) 3

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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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Each shell can hold orbitals equal to $n^2$ , where $n$ is the principal quantum number. For the third shell ( $n=3$ ), the total number of orbitals is 9.

Why orbitals scale with $n^2$

An orbital is a region of space where an electron is likely to be found, defined by three quantum numbers: the principal quantum number $n$ , the azimuthal quantum number $\ell$ , and the magnetic quantum number $m_\ell$ . The third shell means $n = 3$ .

The key insight is that for a given shell, the number of orbitals depends on how many different combinations of $\ell$ and $m_\ell$ are possible. The azimuthal quantum number $\ell$ can range from $0$ to $n-1$ , and for each value of $\ell$ , the magnetic quantum number $m_\ell$ ranges from $-\ell$ to $+\ell$ , giving $(2\ell + 1)$ orbitals per subshell.

Counting orbitals in the third shell

For $n = 3$ , we need to count all possible orbitals across all subshells:

Identify the possible subshells

The azimuthal quantum number $\ell$ can be $0, 1, 2$ (since $\ell$ goes from $0$ to $n-1 = 2$ ).
- $\ell = 0$ corresponds to the $3s$ subshell
- $\ell = 1$ corresponds to the $3p$ subshell
- $\ell = 2$ corresponds to the $3d$ subshell
Count orbitals in each subshell

For each $\ell$ , the number of orbitals is $(2\ell + 1)$ :

Subshell $\ell$ Number of orbitals $(2\ell + 1)$
$3s$ $0$ $2(0) + 1 = 1$

Subshell	$\ell$	Number of orbitals $(2\ell + 1)$
$3s$	$0$	$2(0) + 1 = 1$

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Q.Total number of orbitals associated with third shell will be __________.

Why orbitals scale with n2n^2n2

Counting orbitals in the third shell

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Why orbitals scale with $n^2$

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