Calculate the maximum order of diffraction if X-rays of wavelength 0.819 $A^0$ is incident on a crystal of lattice spacing 0.282 nm. -

Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 8M

Year: Dec 2013

Notations:-

C - Crystal

Co - Collimator

T - Turn Table

Ic - Ionization Chamber

θ - Glancing Angle

Construction:-

• Bragg's Spectrometer consists of a collimator containing two slits $S_1$ and $S_2$ made up of lead, through which X-ray is passed.

• A turn table is situated in-front of the collimator on which crystal is placed.

• Ionization chamber collects the reflected X-ray

Procedure:-

• A fine beam of a monochromatic X-ray is made to fall on the crystal.

• The crystal reflects the X-rays which are collected by the ionization chamber.

• Turn table is rotated till a sharp increase in the intensity is detected.

• The sudden increase in intensity suggests that Bragg's Law is satisfied at the given angle θ.

• Then the inter-planar spacing can be determined by using Bragg's Law nλ = 2d.sinθ

• The peak in ionization current occurs more than once for different values of "n"

Numerical

Given:-

λ = 0.819 $A^0$ = 0.0819 nm

d = 0.282 nm

To Find:-

Maximum order of Diffraction $(n_{Max})$

Solution:-

The maximum order of diffraction will occur when sin θ = 1

Therefore, from Bragg's Law, we can write that;

$n_{Max} λ = 2d$

$ ∴ n_{Max} = \frac{2d}{λ}$

$ ∴ n_{Max} = \frac{(2 × 0.282)}{0.0819}$

$ ∴ n_{Max} = 6.886$

Hence, we will obtain 6th order maxima and not 7th order maxima

$ ∴ n_{Max} = 6$