Hydrogen atom has only one electron, so mutual repulsion between electrons is absent. However, in multielectro

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Q. Hydrogen atom has only one electron, so mutual repulsion between electrons is absent. However, in multielectron atoms mutual repulsion between the electrons is significant. How does this affect the energy of an electron in the orbitals of the same principal quantum number in multielectron atoms?

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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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In multi-electron atoms, electron-electron repulsion causes shielding, which reduces the effective nuclear charge experienced by outer electrons. Different subshells ( $s, p, d, f$ ) penetrate the electron cloud to varying degrees, leading to different effective nuclear charges and thus splitting the energy levels within the same principal quantum number, with the energy order being $s < p < d < f$ .

In a hydrogen atom, which has only one electron, the energy of an electron depends solely on its principal quantum number, $n$ . This means that all orbitals with the same $n$ (e.g., $2s$ and $2p$ ) have the same energy; they are degenerate. However, the situation changes dramatically in multi-electron atoms due to the presence of multiple electrons.

The core concept here is that the mutual repulsion between electrons significantly alters the potential energy landscape within the atom. This repulsion leads to two interconnected phenomena: shielding and penetration, which ultimately determine the energy of an electron in a given orbital.

Here's how it affects the energy of an electron in orbitals of the same principal quantum number:

Hydrogen vs. Multi-electron Atoms:

In a hydrogen atom, the electron experiences the full nuclear charge ( $+Z$ ). Its energy is given by $E_n = -13.6 \frac{Z^2}{n^2} \text{ eV}$ . Since there's only one electron, there's no electron-electron repulsion or shielding.

In multi-electron atoms, each electron experiences not only the attraction from the nucleus but also repulsion from all other electrons.
Electron-Electron Repulsion and Shielding:

The mutual repulsion between electrons means that an electron in an outer orbital is repelled by electrons in inner orbitals (and to some extent, by other electrons in the same principal shell). This repulsion effectively "shields" or "screens" the outer electron from the full positive charge of the nucleus. It's as if the inner electrons are partially canceling out the nuclear charge.
Effective Nuclear Charge ( $Z_{eff}$ ):

Due to shielding, an electron in a multi-electron atom does not experience the full nuclear charge $Z$ . Instead, it experiences a reduced charge, called the effective nuclear charge, $Z_{eff}$ .
$Z_{eff} = Z - S$
Penetration Effect:

Orbitals with the same principal quantum number $n$ but different azimuthal quantum numbers $l$ (i.e., different subshells like $s, p, d, f$ ) have different shapes and radial probability distributions. This means they spend different amounts of time closer to the nucleus. This ability to "penetrate" closer to the nucleus is called the penetration effect.
- An $s$ orbital (for a given $n$ ) has a higher probability of being found very close to the nucleus compared to a $p$ orbital.
- A $p$ orbital penetrates more than a $d$ orbital.
- A $d$ orbital penetrates more than an $f$ orbital. Therefore, the order of penetration for orbitals within the same principal quantum number is $s > p > d > f$ .
Relating Penetration to Shielding and $Z_{eff}$ :
- $s$ orbitals: Because $s$ orbitals penetrate closer to the nucleus, their electrons experience less shielding from inner electrons. This means they experience a higher $Z_{eff}$ . …

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Q.Hydrogen atom has only one electron, so mutual repulsion between electrons is absent. However, in multielectron atoms mutual repulsion between the electrons is significant. How does this affect the energy of an electron in the orbitals of the same principal quantum number in multielectron atoms?

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