Question:
The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at:
The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at:
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Balmer series:
n₁ = 2, n₂ = 3,4,5…
First line always corresponds to 3 → 2.
Updated On: Mar 19, 2026
- \(\dfrac{9R}{400}\,\text{cm}^{-1}\)
- \(\dfrac{7R}{144}\,\text{cm}^{-1}\)
- \(\dfrac{3R}{4}\,\text{cm}^{-1}\)
- (5R)/(36)cm⁻1
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The Correct Option is D
Solution and Explanation
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))
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