Fyneman exercise I ch31-1

Q.

Find the index of refraction of Aluminum for x-rays of wavelength $1.56\times 10^{-8} cm$. Assume that all of the electrons in Aluminum have natural frequencies very much less than the frequency of the x-rays.

A.

Index of refraction eq.(31-19)

$n=1+\frac{Nq_e^2}{2\epsilon_0m(\omega_0^2-\omega^2)}$

symbol	meaning	value
m	mass of electron	$9.11\times10^{-31}$ kilograms
$q_e$	charge of electron	$1.60\times10^{-19}$coulombs
$\epsilon_0$	Vacuum permittivity	$8.85\times10^{-12}$ F/m
$N$	Number of charges per unit volume	see below.

Aluminum properties
molar mass	$26.98 \ g/mol$
density	$2.70 \ g/cm^3$
atomic number	$13$

Ref
$N_A$	Avogadro constant	$6.02 \times 10^{23} mol^{-1}$

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number of charges in unit volume

Number of charges per unit volume can be thought

$N=number\ of\ Aluminum\ atoms\ in\ unit\ volume$
and
$N = Avogadro\ constant \times \frac{density}{ molar\ mass}$
where,

density = $2.70\ g/cm^3 = 2.70 \times 10^{3}\ kg/m^3$
molar mass = $26.98 \times 10^{-3}\ kg/mol$

$\therefore N=6.02 \times 10^{28} \ \# \ of \ charges/m^3$

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Angular frequency $\omega$

$wavelength=period \times velocity$

$\lambda=\frac{2\pi}{\omega}\times c$

$\therefore \omega = \frac{2\pi \times 3.00 \times 10^8}{1.56\times 10^{-10} }=4\pi \times 10^{18}$

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resonance frequency $\omega_0$ is negligible.

Problem state

natural frequencies very much less than the frequency of the x-rays

So, index of refraction is

$n=1-\frac{Nq_e^2}{2\epsilon_0m\omega^2}$

result

$n=1-\frac{6.02 \times 10^{28} \times (1.60\times10^{-19})^2 }{ 2\times 8.85\times10^{-12} \times 9.11\times10^{-31} \times (4\pi \times 10^{18})^2 }=0.999999394$