The wavelength of the first line of the Balmer series in a hydrogen atom is identical to the wavelength of a spectral line emitted during a transition from n=x to n=y (where x > y ) in a Li ²⁺ ion. The values of x and y are respectively

Step-by-step solution

For the first line of the Balmer series in a hydrogen atom ( Z=1 ), the transition is from n=3 to n=2 . The wave number is given by: 1 _H = R_H (1)^2 ( 1 2^2 - 1 3^2 ) = R_H ( 1 4 - 1 9 ) = R_H ( 5 36 ) For the Li ²⁺ ion ( Z=3 ), the transition is from n=x to n=y . The wave number is: 1 _ Li = R_H (3)^2 ( 1 y^2 - 1 x^2 ) = 9 R_H ( 1 y^2 - 1 x^2 ) Given that the wavelengths are equal, _H = _ Li , so their wave numbers are also equal: R_H ( 5 36 ) = 9 R_H ( 1 y^2 - 1 x^2 ) 5 324 = 1 y^2 - 1 x^2 We can rewrite 5 324 a