The wavelength of the radiation emitted when a hydrogen atom electron falls from infinity to 2nd excited state would be: (Rydberg constant =1.097×107m−1)
A
406 nm
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B
192 nm
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C
821 nm
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D
168 nm
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Solution
The correct option is C 821 nm Second excited state means n=3 Using the Rydberg formula, 1λ=RH(1n21−1n22) putting up the values, we get 1λ=1.097×107(132−1∞)=1.218×106λ=821.02×10−9m≈821nm