If the wavelength of the first line of the Balmer series of hydrogen is `6561 Å`, the wavelngth of the second line of the series should be

Options

A. 3575 Å
B. 3860 Å
C. 4500 Å
D. 4860 Å

Answer

D

Explanation

For first line of Balmer series,
`p=2` and `n=3`
`1/lambda_(1)=R(1/p^(2)-1/n^(2))=R(1/4-1/9)`
`:. lambda_(1)=36/(5R)` ...(1)
For second line of Balmer series,
`p=2` and `n=4`
`1/lambda_(2)=R(1/p^(2)-1/n^(2))=R(1/4-1/16)`
`:. lambda_(2)=64/(12R)` ...(2)
Dividing eq. (2) by eq. (1) we have,
`lambda_(2)=(64xx5xxlambda_(1))/(12xx36)`
`=(64xx5xx6561)/(12xx36)=4860 Å`