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 :

Answer

4

Explanation

form `1/(lambda)=RZ^(2)[1/(n_(1)^(2))-1/(n_(2)^(2))]`
for hydrogen, Z=1, for visible radiation,
`n_(2)=n , n_(1)=2`
`1/lambda=R(1/(2^(2))-1/(n^(2)))=R((n^(2)-4))/(4n^(2))`
`lambda=(4n^(2))/(R(n^(2)-4))=(kn^(2))/(n^(2)-4)`, were
`k=4//R=m/R`, where `m=4`