Chapter 25, Problem 55P

To determine

Part(a)To determine:

The angle at which the second order Bragg’s diffraction is observed.

Answer to Problem 55P

Solution:

The second order diffraction maximum is observed at the angle 53.8^o.

Explanation of Solution

X rays are electromagnetic waves of very short wavelengths. When X rays are directed on to the surface of a crystal, they are diffracted since the crystal behaves as a three dimensional diffraction grating. The diffracted beams produce maxima and minima depending on their path difference.

If the spacing between the planes of a crystal is d and the X rays of wavelength $\lambda$ are incident on the planes at a glancing angle $\varphi$, then according to Bragg’s equation, the m^th maximum is obtained according to the expression,

$m\lambda =2d\sin \varphi$

Write the expression for the first and the second order maxima giving values of 1 and 2 to m

$\begin{array}{c}\lambda =2d\sin {\varphi }_1\\ 2\lambda =2d\sin {\varphi }_2\end{array}$

Therefore,

$\frac{\sin {\varphi }_2}{\sin {\varphi }_1}=2$

The glancing angle for the second order maximum is obtained from the expression,

$\begin{array}{c}\sin {\varphi }_2=2\sin {\varphi }_1\\ {\varphi }_2={\sin }^{-1}\left(2\sin {\varphi }_1\right)\end{array}$

Given:

The glancing angle for the first order maximum ${\varphi }_1=23.8°$

The plane separation in the crystal $d=0.24\text{ nm}$

Formula:

The glancing angle for the second order maximum is calculated using,

${\varphi }_2={\sin }^{-1}\left(2\sin {\varphi }_1\right)$

Calculation:

Substitute the value of the glancing angle for first order maximum in the expression and simplify.

$\begin{array}{c}{\varphi }_2={\sin }^{-1}\left(2\sin {\varphi }_1\right)\\ ={\sin }^{-1}\left[2\sin \left(23.8°\right)\right]\\ ={\sin }^{-1}\left(0.8070\right)\\ =53.8°\end{array}$

To determine

Part (b)To determine:

The wavelength of the X rays.

Answer to Problem 55P

Solution:

The wavelength of the given X rays is found to be 0.19 nm.

Explanation of Solution

The wavelength of X rays id calculated using Bragg’s law of diffraction,

$m\lambda =2d\sin \varphi$

For the first order maximum,

$\lambda =2d\sin {\varphi }_1$

Given:

The glancing angle for the first order maximum ${\varphi }_1=23.8°$

The plane separation in the crystal $d=0.24\text{ nm}$

Formula:

$\lambda =2d\sin {\varphi }_1$

Calculation:

Substitute the value of the glancing angle for first order maximum and the plane separation in the expression and simplify.

$\begin{array}{c}\lambda =2d\sin {\varphi }_1\\ =2\left(0.24\text{ nm}\right)\left[\sin \left(23.8°\right)\right]\\ =0.19\text{ nm}\end{array}$