Question

Question: The distance between members of a set of equally spaced planes of atoms in crystalline sodium is $\mathbf{4}\mathbf{.}\mathbf{28}\stackrel{\mathbf{\circ }}{\mathbf{A}}$. If X-rays with $\mathit{\lambda }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{539}\stackrel{\mathbf{\circ }}{\mathbf{A}}$are diffracted by this set of parallel planes, calculate the angle $\mathbf{2}\mathit{\theta }$at which second-order Bragg diffraction will be observed.

Short answer

The angle $2\theta$at which second-order Bragg diffraction will be observed is $42°$.

Step-by-step solution

The Bragg diffraction of x-rays.

Bragg’s equation is$2d\sin \theta =n\lambda$

Here,

d is the distance between the scattering planes in the crystal.

$\theta$is the angle of diffraction.

n is the order of diffraction.

$\lambda$is the wavelength of X-rays.

Given;

The second-order Bragg diffraction (n=2).

$\lambda =1.539\stackrel{\circ }{A}$

$d=4.28\stackrel{\circ }{A}$

The angle at which second-order Bragg diffraction is observed.

As,

$2d\sin \theta =n\lambda$

Therefore,

$\theta =\frac{n\lambda }{2\times \sin \times d}$

Putting the given values in the above equation,

$\theta =\frac{2\times 1.539{\displaystyle \stackrel{\circ }{A}}}{2\times \sin \times 4.28{\displaystyle \stackrel{\circ }{A}}}$

$\theta ={\sin }^{-1}0.36$

Also, ${\sin }^{-1}0.36=21°$, therefore;

$\theta =21°$

And,

$2\theta =42°$