The wavelength of the first spectral line in the Balmer series of hydrogen atom is `6561 A^(@)` . The wavelength of the second spectral line in the Balmer series of singly - ionized helium atom isA. `1215A^(0)`B. `1640A^(0)`C. `2430A^(0)`D. `4687A^(0)`

The wavelength of the first spectral line in the Balmer series of hydr

1.	The wavelength of the first spectral line in the Balmer series of hydrogen atom is `6561 A^(@)` . The wavelength of the second spectral line in the Balmer series of singly - ionized helium atom isA. `1215A^(0)`B. `1640A^(0)`C. `2430A^(0)`D. `4687A^(0)`
Answer» Correct Answer - A `1/(6561)=R(1/4-1/9)=(5R)/(36)(therefore1/lambda=RZ^(2)(1/n_(1)^(2)-1/n_(2)^(2)))` ` 1/lambda=4R(1/4-1/16)=(3Rxx4)/16Rightarrowlambda1215A^(0)`

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The wavelength of the first spectral line in the Balmer series of hydrogen atom is `6561 A^(@)` . The wavelength of the second spectral line in the Balmer series of singly - ionized helium atom isA. `1215A^(0)`B. `1640A^(0)`C. `2430A^(0)`D. `4687A^(0)`

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