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 is

Options

A. `1215A^(0)`
B. `1640A^(0)`
C. `2430A^(0)`
D. `4687A^(0)`

Answer

A

Explanation

`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)`