What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the groun

Exercises · 2.15

Q. What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?

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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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When an electron cascades from $n=6$ to $n=1$ , it can take multiple paths, emitting photons for every downward transition. The total number of unique spectral lines equals all possible jumps between levels: $\frac{n(n-1)}{2} = 15$ .

Why emission lines multiply: the cascade principle

When a hydrogen atom's electron sits in an excited state, it doesn't have to fall directly to the ground state in one leap. Quantum mechanics permits any downward transition— $n=6 \to 5$ , or $6 \to 3$ , or $5 \to 2$ , and so on. Each jump releases a photon whose energy (and thus wavelength) is determined by the energy difference between those two specific levels.

The question asks for the maximum number of lines. That means we want to count every distinct transition that could possibly occur as the electron works its way down from $n=6$ to $n=1$ , across all possible paths taken by all the atoms in a sample.

Counting all possible transitions

The key insight: each emission line corresponds to a unique ordered pair of levels $(n_i, n_f)$ where $n_i > n_f$ .

Identify the levels involved.

The electron starts at $n=6$ and ends at $n=1$ , so the accessible levels are $n = 1, 2, 3, 4, 5, 6$ .
Count the downward jumps.

A transition can occur from any higher level to any lower level. Mathematically, we're choosing 2 distinct levels from the 6 available and assigning the higher one as the initial state. This is a combination problem:

\text{Number of lines} = \binom{6}{2} = \frac{6 \times 5}{2} = 15

Enumerate them systematically (optional verification). From $n=6$ : five transitions ( $6 \to 5, 6 \to 4, 6 \to 3, 6 \to 2, 6 \to 1$ ) From $n=5$ : four transitions ( $5 \to 4, 5 \to 3, 5 \to 2, 5 \to 1$ ) …

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Q.What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?

Why emission lines multiply: the cascade principle

Counting all possible transitions

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