For the first member of Balmer series of hydrogen spectrum, the wavelength is `lamda`. What is the wavelength of the second member?A. `(20lambda)/(27)`B. `(3lambda)/(16)`C. `(5lambda)/(36)`D. `(3lambda)/(4)`

For the first member of Balmer series of hydrogen spectrum, the wavele

1.	For the first member of Balmer series of hydrogen spectrum, the wavelength is `lamda`. What is the wavelength of the second member?A. `(20lambda)/(27)`B. `(3lambda)/(16)`C. `(5lambda)/(36)`D. `(3lambda)/(4)`
Answer» Correct Answer - A `(1)/(lambda_(1))=R ((1)/(4)-(1)/(9)) rArrlambda_(1)=(4xx9)/(5R)` Similarly `(1)/(lambda_(2))=R((1)/(4)-(1)/(4^(2))) rArrlambda_(2)=(16)/(3R)=(16)/(3)xx(5lambda)/(4xx9)=(20)/(27)lambda`

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For the first member of Balmer series of hydrogen spectrum, the wavelength is `lamda`. What is the wavelength of the second member?A. `(20lambda)/(27)`B. `(3lambda)/(16)`C. `(5lambda)/(36)`D. `(3lambda)/(4)`

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