Eₙ = -13.6/n² eV. Calculate wavelength of Balmer series transition.

Bohr Model — Energy Levels of Hydrogen

Question

An electron in a hydrogen atom transitions from the $n = 4$ level to the $n = 2$ level. Calculate the wavelength of the emitted photon. (Given: $R_H = 1.097 \times 10^7 \text{ m}^{-1}$ )

This is a classic Balmer series problem — it appeared in JEE Main 2023 and shows up every year in CBSE board exams with minor variations.

Solution — Step by Step

We’re going from $n_i = 4$ to $n_f = 2$ . Since $n_f = 2$ , this belongs to the Balmer series — the only hydrogen series that falls in the visible light range.

Knowing this upfront helps us sanity-check our answer: Balmer wavelengths lie between ~380 nm and ~656 nm.

The Rydberg formula gives us the wavenumber (inverse wavelength) of any hydrogen spectral line:

\frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)

Here $R_H = 1.097 \times 10^7 \text{ m}^{-1}$ is the Rydberg constant. The formula always puts the lower level in the first term — this ensures we get a positive value.

\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{4^2} \right)

\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{16} \right)

\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{3}{16}

\frac{1}{\lambda} = 1.097 \times 10^7 \times 0.1875 = 2.057 \times 10^6 \text{ m}^{-1}

\lambda = \frac{1}{2.057 \times 10^6} \approx 4.86 \times 10^{-7} \text{ m}

\boxed{\lambda \approx 486 \text{ nm}}

This is the H-beta line — the blue-green line visible in hydrogen’s emission spectrum.

Why This Works

The Bohr model quantises electron energy at level $n$ as $E_n = -\frac{13.6}{n^2}$ eV. When an electron drops from a higher level to a lower one, it releases energy as a photon. The energy of that photon equals exactly the difference between the two levels.

For $n=4$ to $n=2$ : the energy released is $E_4 - E_2 = -0.85 - (-3.4) = 2.55$ eV. Using $E = hc/\lambda$ , this gives $\lambda \approx 486$ nm — consistent with what the Rydberg formula gives us. Both routes lead to the same answer.

The Rydberg formula is essentially a compact, pre-derived version of this energy difference calculation. In exam conditions, using Rydberg directly saves about 40 seconds over the $E_n$ route.

Alternative Method

We can use energy levels directly instead of Rydberg.

E_4 = \frac{-13.6}{16} = -0.85 \text{ eV}, \quad E_2 = \frac{-13.6}{4} = -3.4 \text{ eV}

Energy of emitted photon:

\Delta E = E_4 - E_2 = -0.85 - (-3.4) = 2.55 \text{ eV}

Convert to wavelength using $\lambda = \frac{hc}{\Delta E}$ :

\lambda = \frac{1240 \text{ eV·nm}}{2.55 \text{ eV}} \approx 486 \text{ nm}

Memorise $hc = 1240$ eV·nm. This lets you skip unit conversions entirely when energy is in eV and you want wavelength in nm. This trick alone saves marks in JEE Main’s time-crunched conditions.

Common Mistake

Students frequently write the Rydberg formula as $\frac{1}{\lambda} = R_H \left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right)$ — putting the initial level first. For an $n=4 \to n=2$ transition, this gives $\frac{1}{16} - \frac{1}{4}$ , which is negative. A negative wavelength should immediately signal an error. The formula always has the lower (final) level in the first fraction so the result stays positive. The electron loses energy, so $n_f < n_i$ — the final level’s term must be larger.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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