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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`. |
Answer» 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)` |
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