Question

If the series limit wavelength of the Lyman series for hydrogen atom is 912AA, then the series limit wavelength for the Balmer series of hydrogen atom is:

• A. \(912\,\text{\AA}\)
• B. \(912\times2\,\text{\AA}\)
• C. \(912\times4\,\text{\AA}\)
• D. (912)/(2)AA

Hint:

Balmer series wavelengths are four times those of Lyman series.

Answer 1

The Correct Option is C

Step 1: Lyman series limit corresponds to transition n=∞ → 1.
Step 2: Balmer series limit corresponds to transition n=∞ → 2.
Step 3: Since wavelength λ ∝ n²: lambdaBₐlmₑᵣ = 4lambdaLyman = 912×4AA