Question

An x-ray beam of a certain wavelength is incident on an NaCl crystal, at $\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{°}$to a certain family of reflecting planes of spacing$\mathbf{39}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{pm}\mathbf{.}$. If the reflection from those planes is of the first order, what is the wavelength of the x rays?

Short answer

The wavelength of the x rays will be ${\lambda }_A=39.8 \text{pm}$.

Step-by-step solution

Introduction of diffraction

Diffraction is a phenomena in which the interference or the bending of waves occurs around the corners of the opening or obstacle through or on which it strikes.

The diffraction formula is following,

$$\begin{gathered} \mathrm{n\lambda }=2\mathrm{dsin\theta } \\ \mathrm{\lambda }=\text{wavelength} \\ \mathrm{d}=\text{spacing} \\ \mathrm{\theta }=\text{bragg angle} \\ \mathrm{n}=\text{order of diffraction} \end{gathered}$$

Step 2: Identification of the given data

The reflecting planes of spacing is,$d=39.8 \text{pm}$ .

The x-ray incident is, $\theta =30.0°$.

The wavelength of the X rays

The wavelength of the x ray will be following with the help of given data and formula,

$\lambda =2d\sin \theta$

Substitute all the value in the above equation.

$$\begin{gathered} {\lambda }_A=2\left(39.8 \text{pm}\right)\sin 30° \\ {\lambda }_A=39.8 \text{pm} \end{gathered}$$

Hence the wavelength of the x rays will be ${\lambda }_A=39.8 \text{pm}$.