A brag diffraction experiment is conducted using a beam of electrons accelerated through a$\mathbf{\text{1.0\hspace{0.17em}kV}}$potential difference. (a) If the spacing between atomic planes in the crystal is$\mathbf{\text{0.1nm}}$, at what angles with respect to the planes will diffraction maximum be observed? (b) If a beam of$\mathbf{X}$-rays products diffraction maxima at the same angles as the electron beam, what is the$\mathbf{X}$-ray photon energy?

So, we'll use the relationship shown in Example $\mathbf{\text{4.3}}$directly, linking the particle's wavelength $\mathbf{\text{(}}\mathit{\lambda }\mathbf{\text{)}}$to the potential for rapid growth (V). can use Bragg's law to find the equivalent values of once have the wavelength, such that,

$\mathit{\lambda }\mathbf{\text{=}}\sqrt{\frac{{\mathbf{\text{h}}}^{\mathbf{\text{2}}}}{\mathbf{\text{2mqV}}}}$………………..(1)

Substitute the given data in equation (1), and we get,

$\begin{array}{c}\lambda =\sqrt{\text{1.5}\times {\text{10}}^{\text{-21}}{\text{m}}^{\text{2}}}\\ =\text{3.89}\times {\text{10}}^{\text{-11}}\text{m}\end{array}$

$\begin{array}{c}\text{2dsin(θ)}=\text{n}\lambda \\ \text{sin(θ)}=\left(\frac{\lambda }{\text{2d}}\right)\text{n}\\ \text{sin(θ)}=\left(\frac{\text{3.89}\times {\text{10}}^{\text{-11}}\text{m}}{\text{2}\times \text{0.1}\times {\text{10}}^{\text{-9}}\text{m}}\right)\times \text{n}\\ \text{θ}={\text{sin}}^{\text{-1}}\left(\text{0.194n}\right)\end{array}$

$\begin{array}{l}{\text{θ}}_{\text{1}}=\text{11.2}°\\ {\text{θ}}_{\text{2}}=\text{22.83}°\\ {\text{θ}}_{\text{3}}=\text{35.59}°\\ {\text{θ}}_{\text{4}}=\text{50.89}°\\ {\text{θ}}_{\text{5}}=\text{75.93}°\end{array}$

Keep in mind that the value of$\text{n=6}$ isn't allowed because arcsine's parameter exceeds$\text{1}$ , but used the integer n instead of m for the integer counter so it doesn't get mixed up with the unit of measurement ($\text{m}$ or meter).

(b)

To achieve the same maxima, both the electron and the photons must have the same wavelength value.

${\text{E}}_{\text{p}}\text{=}\frac{\text{hc}}{\lambda }$

$\begin{array}{c}{\text{E}}_{\text{p}}=\left(\frac{4.14\times {10}^{-23}eV\cdot s\times \text{3}\times {\text{10}}^{\text{8}}\text{\hspace{0.17em}m}/s}{\text{3.89}\times {\text{10}}^{\text{-11}}\text{m}}\right)\\ =\text{31.88keV}\end{array}$

Therefore, the energy of the photon is${\text{E}}_{\text{p}}=\text{31.88keV}$ .