The first line in the Lyman series has wavelength `lambda`. The wavelegnth of the first line in Balmer series is

Options

A. `36/5 lamda`
B. `5/36 lamda`
C. `27/5 lamda`
D. `5/27 lamda`

Answer

C

Explanation

For Lyman series `n_f = 1, n_i =2` for the first line
`therefore 1/lamda = R[1/2^2 - 1/2^2 ] = (3R)/(4)`
`therefore lamda=(4)/(3R) " " ..(1)`
For the first line Balmer series , `n_f = 2, n_i =3`
`therefore 1/lamda_1 = R [1/2^2 - 1/3^2] = (5R)/(36)`
`therefore lamda_1 = (36)/(5R) " "...(2)`
` therefore lamda_1/lamda_2 = (36)/(5R) xx (3R)/(4) = 27/5 therefore lamda_1 = 27/5 lamda`