Atomic Scattering Factor

Atomic Scattering Factor

The ratio of amplitude of wave scattered from the atom or certain charge distribution to the amplitude of the wave scattered from the single electron located at a point.

Let us consider a spherically symmetric charge distribution of the atom having an electron at it’s centre as shown in the following figure below.

FIGURE HERE : Atomic scattering of X- ray from point electron at O and from the charge distribution

Let us consider \(\rho(r)\) be the volume charge density then the charge distribution at a distance r from the centre O within the volume element d\(\tau\) is \(\rho(r)d\tau\).

Let \(n_1\) be the direction of incident wave which is scattered from the point electron at O having scattering amplitude \(e^{i(kx-\omega t)}\) ; x is the distance covered along \(n_2\) .

Then radiation scattered from the charge distribution \(\rho(r)d\tau\) is always in phase difference with the radiation scattered from point electron; And this phase difference is given by $$\phi_r=\frac{2\pi}{\lambda}(\vec r. \vec N)$$

Where \(\vec N\) is a unit vector, and value of \(\vec N = 2sin\theta\)

Now, the amplitude of the wave scattered from the charge distribution \(\rho(r)d\tau\) is given by $$e^{i(kx-\omega t+\phi_r)}\rho(r)d\tau$$

Atomic scattering factor is given by $$df=\frac{e^{i(kx-\omega t+\phi_r)}\rho(r)d\tau}{e^{i(kx-\omega t)}}$$

$$df=e^{i\phi_r}\rho(r)d\tau$$

Now, the ratio atomic scattering factor due to whole charge distribution is given by

$$f=\int_v df=\int_v e^{i\phi_r}\rho(r)d\tau\dotsm(1)$$

Now, \(\phi_r=\frac{2\pi}{\lambda}\vec r. \vec N=\frac{2\pi}{\lambda}rNcos\phi\)

$$=\frac{2\pi}{\lambda}rcos\phi 2sin\theta=\frac{4\pi}{\lambda}sin\theta r cos\phi$$

$$\therefore\;\;\phi_r=\mu r cos\phi$$

Where \(\mu=\frac{4\pi}{\lambda}sin\theta\)

and volume element in spherical polar co-ordinate system is \(d\tau=2\pi r^2sin\phi d\phi dr\)

therefore equation (1) becomes;

$$f=\int_v e^{i\mu r cos\phi}\rho(r).2\pi r^2 sin\phi d\phi dr$$

$$f=\int^\infty_0\int^\pi_0 e^{i\mu rcos\phi}\rho(r) 2\pi r^2sin\phi d\phi dr\dotsm(2)$$

Consider $$I=\int^\pi_0 e^{i\mu r cos\phi}sin\phi d\phi$$

Put \(cos\phi = x\)

So, \(-sin\phi d\phi = dx\)

And, \(x_1=1 and x_2=-1\)

Then, $$I=-\int^-1_1 e^{i\mu r x}dx$$

$$I=\int^1_-1 e^{i\mu rx}dx$$

$$=\biggl[\frac{e^{i\mu rx}}{i\mu r}\biggr]^1_{-1}$$

$$=2\biggl[\frac{e^{i\mu r}-e^{-i\mu r}}{2i\mu r}\biggr]$$

$$I=2\frac{sin\mu r}{\mu r}$$

Therefore equation (2) becomes;

$$f=\int^\infty_0 2\pi r^2 \rho(r)\frac{2sin\mu r}{\mu r}dr$$

This is the required expression for atomic scattering factor.

References:

Adhikari, P.B, Daya Nidhi Chhatkuli and Iswar Prasad Koirala.A Textbook of Physics. Vol. II. Kathmandu: Sukunda Pustak Bhawan, 2012.

Jenkins, F.A and H.E White.Fundamental of optics. New York (USA): McGraw-Hill Book Co, 1976.

wood, R.W.Physical Optics. New York (USA): Dover Publication , 1934.