The wavelength of the third line of the Balmer series for a hydrogen atom is :
A
21100RH
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B
10021RH
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C
21RH1000
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D
1000RH21
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Solution
The correct option is B10021RH By using the relation, 1λ=RH(1n21−1n22) for the H-atom. For the Balmer series, n1=2 and n2=5 (for the third line) 1λ=RH(122−152)=21100RH∴λ=10021RH