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

Options

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

Answer

d

Explanation

`(1)/(lambda)=Rz^(2)((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))`
For `H-"atom" z=1`
For visible radiation, `n_(1)=2`
`(1)/(lambda)=R((1)/(2^(2))-(1)/(n^(2)))=(R(n^(2)-4))/(4n^(2))`
but `lambda^(2)=(4x^(2))/(R(n^(2)-4))-(kn^(2))/(n^(2)-4)`
`k=(4)/(R )`