How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with

Exercises · 2.14

Q. How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n = 1 orbit).

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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 energy required to ionise a hydrogen atom from the $n=5$ orbit is $0.544 \times 10^{-19} \, \text{J}$ (or $0.544 \, \text{aJ}$ ). This is only $1/25$ of the ionisation enthalpy from the ground state ( $n=1$ ), which is $2.18 \times 10^{-18} \, \text{J}$ — because the electron is already much less tightly bound in a higher orbit.

The key idea here is energy level quantisation in the Bohr model of the hydrogen atom. The electron can only occupy certain allowed orbits, each with a specific energy. The energy of the electron in the $n$ th orbit is given by:

E_n = -\frac{2.18 \times 10^{-18}}{n^2} \, \text{J}

The negative sign means the electron is bound to the nucleus — you need to supply positive energy to remove it (ionise it). The magnitude of this energy is the ionisation energy from that level.

To ionise the atom, you must give the electron enough energy to escape completely, i.e., raise its total energy from $E_n$ to zero (the free electron state). So the required ionisation energy from orbit $n$ is:

\text{Ionisation energy} = 0 - E_n = -E_n = \frac{2.18 \times 10^{-18}}{n^2} \, \text{J}

Now let's apply this.

For the $n=5$ orbit:

E_5 = -\frac{2.18 \times 10^{-18}}{5^2} = -\frac{2.18 \times 10^{-18}}{25} = -8.72 \times 10^{-20} \, \text{J}

So the ionisation energy from $n=5$ is:

\text{IE}_{n=5} = -E_5 = 8.72 \times 10^{-20} \, \text{J}

Tip

A quick way: since $E_n \propto 1/n^2$ , the ionisation energy from $n=5$ is exactly $1/25$ of the ground state ionisation energy. So you can compute $2.18 \times 10^{-18} / 25 = 8.72 \times 10^{-20} \, \text{J}$ directly.

For the ground state ( $n=1$ ): The ionisation enthalpy of the H atom (from $n=1$ ) is:

\text{IE}_{n=1} = \frac{2.18 \times 10^{-18}}{1^2} = 2.18 \times 10^{-18} \, \text{J}

Comparison: The ratio is: …

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Q.How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n = 1 orbit).

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