The wavelength of second Balmer line in Hydrogen spectrum is 600 nm. The wavelength for its third line in Lymann series is

Options

A. 800 nm
B. 600 nm
C. 400 nm
D. 120 nm

Answer

D

Explanation

we know, for second line in Balmer series,
`1/(lambda_1) =R[1/4-1/(16)]=(3R)/(16)`
For third line is Lymann series,
`1/(lambda_2)=R[1/1-1/(16)]=(15)/(16)R`
`therefore (lambda_2)/(lambda_1)=(3R)/(16)xx(16)/(15R)=1/5`
or, `lambda_2 =1/5xxlambda_1=1/5xx600nm=120 nm`