1. 0.025×10⁴\ m⁻¹ 2. 0.5×10⁷\ m⁻¹ 3. 0.25×10⁷\ m⁻¹ 4. 2.5×10⁷\ m⁻¹

Answer: 3. 0.25×10⁷\ m⁻¹ Key idea: barν=R(dfrac12²-dfrac1n²) ; the last Balmer line has n→∞ . barν=R/4=10⁷/4=0.25×10⁷\ m⁻¹ .

Given the value of Rydberg constant is 10^7\ m^-1, the wave number of the last line of the Balmer series in hydrogen spectrum will be :

1. \(0.025\times10^{4}\ m^{-1}\)

2. \(0.5\times10^{7}\ m^{-1}\)

3. \(0.25\times10^{7}\ m^{-1}\)

4. \(2.5\times10^{7}\ m^{-1}\)

Doubtly Asked 30 Oct 2026 NEET UG 2016 Phase 1 Practice test neet Save WhatsApp Report

Best answer

Answer: 3. \(0.25\times10^{7}\ m^{-1}\)

Key idea: \(\bar\nu=R\left(\dfrac1{2^2}-\dfrac1{n^2}\right)\); the last Balmer line has \(n\to\infty\).

\(\bar\nu=\dfrac R4=\dfrac{10^7}{4}=0.25\times10^{7}\ \text{m}^{-1}\).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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