Chemistry - Atomic Structure Question with Solution | TestHub

ChemistryAtomic StructureSpectrumMedium2 minPYQ_2012

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The limiting line in Balmer series will have a frequency of (Rydberg constant, $R_{\infty} = 3.29 \times 1 0^{15}$ cycles $/ s$ )

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Solution:

$\overset{v}{ˉ} = \frac{1}{λ} = R_{H} Z (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$ In Balmer series $n_{1} = 2& n_{2} = 3, 4, 5 \dots$ ... Last line of the spectrum is called series limit. Limiting line is the line of shortest wavelength and high energy when $n_{2} = \infty$ $∴ \overset{v}{ˉ} = \frac{1}{λ} = \frac{R _{H}}{n _{1}^{2}} = \frac{3.29 \times 1 0 ^{15}}{2 ^{2}} = \frac{3.29 \times 1 0 ^{15}}{4} = 8.22 \times 1 0^{14} s^{- 1}$

Stream:JEESubject:ChemistryTopic:Atomic StructureSubtopic:Spectrum

⏱ 2mℹ️ Source: PYQ_2012

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