Question

The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at:

• A. \(\dfrac{9R}{400}\,\text{cm}^{-1}\)
• B. \(\dfrac{7R}{144}\,\text{cm}^{-1}\)
• C. \(\dfrac{3R}{4}\,\text{cm}^{-1}\)
• D. (5R)/(36)cm⁻1

Hint:

Balmer series: n₁ = 2, n₂ = 3,4,5… First line always corresponds to 3 → 2.

Answer 1

The Correct Option is D

Step 1: Balmer series corresponds to transitions ending at n₁=2.
Step 2: First emission line is for transition: n₂=3 → n₁=2
Step 3: Using Rydberg formula: ν = R((1)/(2²)-(1)/(3²)) = R((1)/(4)-(1)/(9))
Step 4: ν = R((5)/(36))