Question

The binding energies of K-shell and L-shell electrons in copper are 8.979 keV and 0.951 keV, respectively. If a ${\mathbf{K}}_{\mathbf{\alpha }}$x-ray from copper is incident on a sodium chloride crystal and gives a first-order Bragg reflection at an angle of $\mathbf{74}\mathbf{.}\mathbf{1}\mathbf{°}$ measured relative to parallel planes of sodium atoms, what is the spacing between these parallel planes?

Short answer

The spacing between these parallel planes is 80.3 pm .

Step-by-step solution

The given data:

The binding energy of K-shell,${\mathrm{E}}_{\mathrm{k}}=8.979\mathrm{keV}$

The binding energy of L-shell, ${\mathrm{E}}_{\mathrm{L}}=0.951\mathrm{keV}$

The x-ray gives the first-order reflection at an angle $\mathrm{\theta }=74.1°$relative to the parallel planes of sodium atoms.

Consider the known data as below.

The Plank’s constant is,

$$\begin{gathered} \mathrm{h}=6.63\times {10}^{-34}\mathrm{J}.\mathrm{s} \\ =\left(6.242\times {10}^{15}\times 6.63\times {10}^{-34}\right)\mathrm{keV}.\mathrm{s} \\ =41.384\times {10}^{-19}\mathrm{keV}.\mathrm{s} \end{gathered}$$

The speed of light is,

$$\begin{gathered} \mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s} \\ =\left(3\times {10}^8\times {10}^{12}\right)\mathrm{pm}/\mathrm{s} \end{gathered}$$

Understanding the concept of Bragg’s law:

If the X-ray takes place in a crystal area, its incident angle, $\mathit{\theta }$, will show the same scattering angle, θ. Also, when the path difference d is equal to the whole number n of wavelength, $\mathit{\lambda }$, constructive interference will occur.

Using the concept of Bragg's law and the concept of Planck's constant, define the spacing between the parallel planes by considering the energy difference between the K-shell and L-shell binding energies.

Formulas:

According to the Bragg’s law of reflection,

$2\mathrm{dsin\theta }=\mathrm{n\lambda }$ ….. (1)

Here, is the order of reflection, d is the spacing between the two planes of the sodium, and $\lambda$is the wavelength.

The energy of the photon due to Planck’s relation,

$∆\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$ ….. (2)

Here, $∆\mathrm{E}$is the energy of the photon, h is the Plank’s constant, and c is the speed of light

Calculation of the spacing between the parallel planes:

Substituting the given data and equation (2) in equation (1), the spacing between the two planes of the sodium atoms is as follow.

$$\begin{gathered} \mathrm{d}=\frac{\mathrm{nhc}}{2\mathrm{sin\theta }∆\mathrm{E}} \\ \frac{\mathrm{nhc}}{2\mathrm{sin\theta }\left({\mathrm{E}}_{\mathrm{K}}-{\mathrm{E}}_{\mathrm{L}}\right)} \end{gathered}$$

Substitute known values in the above equation.

$$\begin{gathered} \mathrm{d}=\frac{1\times 41.384\times {10}^{-19}\mathrm{keV}.\mathrm{s}\times {10}^{20}\mathrm{pm}/\mathrm{s}}{2\left(\sin 74.1°\right)\left(8.979-0.951\right)\mathrm{keV}} \\ =\frac{1\times 41.384\times {10}^{-19}3\times {10}^{20}}{2\times 0.962\times 8.028}\mathrm{pm} \\ =80.3\mathrm{pm} \end{gathered}$$

Hence, the value of the spacing is 80.3 pm .