The force of repulsion between two electrons kept at a distance of 1 m is F. if m is the mass of the electron, h is the planck's constant and c is the velocity of light, then the Rydberg's constant of

Options

A. `(m pi F)/(h^(3) C)`
B. `(2m pi^(2) F)/(h^(3) C)`
C. `(2m pi^(2) F^(2))/(h^(3) C)`
D. `(m pi F^(2))/(h^(3)C)`

Answer

C

Explanation

`F=1/(4pi in_(0)) e^(2)/r^(2)` but `r=1`
`:. F=1/(4pi in_(0)) e^(2)/1` Now `F^(2)=e^(4)/(16pi^(2) in_(0)^(2))`
Rydberg's constant is given by
`R=(me^(4))/(8c in_(0)^(2) h^(3))`
Multiplying and dividing above eqn by `2 pi^(2)`
`:. R=(me^(4))/(8c in_(0)^(2) h^(3)) (2pi^(2))/(2pi^(2))`
`=(2pi^(2) me^(4))/(h^(3)c16pi^(2) in_(0)^(2)) :' e^(4)/(16 pi^(2) in_(0)^(2))=F^(2)`
`:. R=(2pi^(2) F^(2) m)/(ch^(3)) i.c.R = (2m pi^(2) F^(2))/(ch^(3))`