Wavelength of first line in Lymen series is `lambda`. What is wavelength of first line in Balmer series?

Options

A. `(5lambda)/27`
B. `(27 lambda)/5`
C. `(36 lambda)/5`
D. `(5 lambda)/36`

Answer

B

Explanation

`lambda_(1)=lambda, lambda_(2)=?`
For first of Lyman series `p=1` and `n=2`
`:. 1/lambda_(1)=R[1/p^(2)-1/n^(2)]`
`1/lambda_(1)=R [1/1-1/4]`
`lambda_(1)=4/(3R)` ...(1)
For first line of Balmer `p=2` and `n=3`
`:. 1/lambda_(2)= R[1/4-1/9]`
`lambda_(2)=36/(5R)` ...(2)
`lambda_(2)/lambda_(1)=36/(5R)xx(3R)/4=27/5`
`lambda_(2)=27/5 lambda_(1)=27/5 lambda`