The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes n = 2 → n = 3 and n = 4 → n = 6 transitions, respectively, is

For a hydrogen atom the wave‑number (inverse wavelength) of a photon emitted or absorbed in a transition between levels $n_i$ and $n_f$ is given by the Rydberg formula

$1/\lambda \; =\; R ·���(1/n\_i^2\; –\; 1/n\_f^2)$

Since the Rydberg constant $R$ is the same for all transitions, the wavelength is inversely proportional to the factor

$\Delta \; =\; 1/n\_i^2\; –\; 1/n\_f^2$

Thus for two different transitions the ratio of their wavelengths equals the inverse ratio of the corresponding $Δ$‑values:

$\lambda \_1\; /\; \lambda \_2\; =\; \Delta \_2\; /\; \Delta \_1$

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**Transition $2 → 3$**

Calculate $Δ_{2→3}$:

$\Delta \_\{2\to 3\}\; =\; 1/2^2\; –\; 1/3^2\; =\; 1/4\; –\; 1/9\; =\; (9‑4)/36\; =\; 5/36$

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**Transition $4 → 6$**

Calculate $Δ_{4→6}$:

$\Delta \_\{4\to 6\}\; =\; 1/4^2\; –\; 1/6^2\; =\; 1/16\; –\; 1/36\; =\; (9‑4)/144\; =\; 5/144$

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**Ratio of wavelengths**

Using $λ_1/λ_2 = Δ_{4→6}/Δ_{2→3}$:

$\lambda \_\{2\to 3\}/\lambda \_\{4\to 6\}\; =\; (5/144) ÷ (5/36)\; =\; (1/144) ÷ (1/36)\; =\; 36/144\; =\; 1/4$

Therefore the ratio of the wavelengths for the two given transitions is

$1/4$