Assuming electrons of substance to be free when subjected to hard `X`-rays, determine by what magnitude the refractive index of graphite differs from unity in the case of `X-`rays whose wavelength in vaccume is equal to `lambda = 50 pm`.

Explanation

For hand `x`-rays, the electrons in graphic will behave as if nearly free and the formula of previous problem can be applied. Thus
`n^(2) =1 - (n_(0)e^(2))/(epsilon_(0)m omega^(2))`
and `n = 1 - (n_(0)e^(2))/(2epsilon_(0)m omega^(2))`
on taking square root and neglecting higher order terms.
So `n-1 =- (n_(0)e^(2))/(2epsilon_(0)m omega^(2)) =- (n_(0)e^(2)lambda^(2))/(8pi^(2)epsilon_(0)m e^(2))`
We calacuate `n_(0)` as follows : There are `6 xx 6.023 xx 10^(23)` electrons in `12` gms of graphite of density `1.6 gm//c.c.` Thus
`n_(0) (6 xx 6.023 xx 10^(23))/((12//1.6)) per c.c`
Using the values of other constants and `lambda = 50 xx 10^(-12)` metre we get
`n - 1 =- 5.4 xx 10^(-7)`