The wave number of the first emission line in the Balmer series of H-Spectrum is n 36 R . The value of 'n' is. (R = Rydberg constant):

v ¯ = R Z 2 1 2 2 - 1 3 2 = R 1 4 - 1 9 = 5 R 36 = n 36 R Hence n = 5.

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ChemistryAtomic StructureMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasynumerical

The wave number of the first emission line in the Balmer series of H-Spectrum is $\frac{n}{36} R$ . The value of 'n' is. (R = Rydberg constant):

Answer:

5.00

Solution:

$\bar{v} = R Z^{2} (\frac{1}{2^{2}} - \frac{1}{3^{2}})$

$= R (\frac{1}{4} - \frac{1}{9}) = \frac{5 R}{36} = \frac{n}{36} R$

Hence n = 5.

Stream:NTA_ABHYASSubject:ChemistryTopic:Atomic StructureSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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