For the first member of Balmer series of hydrogen spectrum, the wavelength is `lamda`. What is the wavelength of the second member?

Options

A. `(27)/(20) lambda`
B. `(20)/(27)lambda`
C. `(27)/(10) lambda`
D. None of these

Answer

B

Explanation

The wavelength of series of n is given by
`" " (1)/(lambda)=R((1)/(2^(2))-(1)/(n^(2)))`
where , R is Rhydberg's constant
For Balmer series, n=3 gives the member of series and n=4 gives the second member of series, Hence,
`" " (1)/(lambda)=R((1)/(2^(2))-(1)/(3^(2)))=R((5)/(36)) " " ....(i)`
`rArr" " (1)/(lambda)=R((1)/(2^(2))-(1)/(4^(2)))=R((12)/(16xx4))=(3R)/(16)" " ....(ii)`
`rArr" " (lambda_(2))/(lambda_(2))=(16)/(3)xx(5)/(36)=(20)/(27)`
Given `" " lambda_(1)=lambdarArrlambda_(2)=(20)/(27)lambda`