Question

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

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

Hint:

Series limit wavelength varies as: λ ∝ n₁² Balmer limit is four times the Lyman limit.

Answer 1

The Correct Option is B

Step 1: Rydberg formula: (1)/(λ) = R((1)/(n₁²)-(1)/(n₂²))
Step 2: Series limit corresponds to n₂→∞. For Lyman series (n₁=1): lambdaL = (1)/(R) = 912AA
Step 3: For Balmer series (n₁=2): lambdaB = (1)/(R((1)/(4))) = 4lambdaL
Step 4: lambdaB = 4×912 = 3648AA