What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n =

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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 Balmer transition $n=4 \to n=2$ in $\text{He}^+$ has the same wavelength as the $n=2 \to n=1$ (Lyman-alpha) transition in hydrogen, because the energy scales with $Z^2$ and the level-difference factor happens to match.

Why wavelengths can match across different atoms

The energy of a photon emitted during a transition depends on two things: the nuclear charge (which sets the overall energy scale) and the specific quantum levels involved. For hydrogen-like ions, the energy of level $n$ is

E_n = -13.6 \, \text{eV} \cdot \frac{Z^2}{n^2}

where $Z$ is the atomic number. Hydrogen has $Z=1$ , while $\text{He}^+$ (singly ionized helium) has $Z=2$ . The factor of $Z^2 = 4$ makes all energy levels in $\text{He}^+$ four times deeper than in hydrogen.

When an electron transitions from level $n_i$ to $n_f$ , the photon energy is

\Delta E = 13.6 \, Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \, \text{eV}

Since wavelength $\lambda = \frac{hc}{\Delta E}$ , two transitions have the same wavelength when their $\Delta E$ values are equal—even if they occur in different atoms.

Finding the matching hydrogen transition

1. Calculate the energy released in the $\text{He}^+$ transition $n=4 \to n=2$

For $\text{He}^+$ , $Z=2$ :

\Delta E_{\text{He}^+} = 13.6 \times 4 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 54.4 \left( \frac{1}{4} - \frac{1}{16} \right)

= 54.4 \left( \frac{4-1}{16} \right) = 54.4 \times \frac{3}{16} = 10.2 \, \text{eV}

2. Set up the equation for hydrogen ( $Z=1$ )

We need a transition in hydrogen with the same energy:

13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) = 10.2

\frac{1}{n_f^2} - \frac{1}{n_i^2} = \frac{10.2}{13.6} = \frac{3}{4}

3. Solve for the quantum numbers

We need integer values of $n_f$ and $n_i$ (with $n_i > n_f$ ) such that:

\frac{1}{n_f^2} - \frac{1}{n_i^2} = 0.75

Try $n_f = 1$ :

1 - \frac{1}{n_i^2} = \frac{3}{4}

\frac{1}{n_i^2} = \frac{1}{4} \implies n_i = 2

This works! The transition is $n=2 \to n=1$ in hydrogen. …

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Q.What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+He^+He+ spectrum?

Why wavelengths can match across different atoms

Finding the matching hydrogen transition

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Q.What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of $He^+$ spectrum?

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