Balmer gave an equation for wavelength of visible radiation of H-spectrum as `lambda=(kn^(2))/(n^(2)-4)`. The value of k in terms of Rydberg's constant R is `m//R`, where m is :

Options

A. R
B. 4R
C. R/4
D. 4/R

Answer

D

Explanation

As, `(1)/(lamda)=RZ^(2)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
For H-atom Z=1, for visible
radiation, n=2
`(1)/(lamda)=R[(1)/(2^(2))-(1)/(n^(2))]=(R(n^(2)-4))/(4n^(2))`
Given `lamda=(kn^(2))/(n^(2)-4)impliesk=(4)/(R)`