Question

A beam of electrons of energy $\mathrm{E}$ scatters from a target having atomic spacing $1\stackrel{\circ }{\mathrm{A}}.$The first maximum intensity occurs at $\mathrm{\theta }=60°$ Then $\mathrm{E}$ (in eV) is __________.

(Planck constant $6.64\times {10}^{-34}\mathrm{js},1\mathrm{eV}=1.6\times {10}^{-19}\mathrm{Js},1\mathrm{eV}=1.6\times {10}^{-19}\mathrm{J},$electron mass m=$9.1\times {10}^{-31}\mathrm{kg}$)

Answer 1

Answer: $50\mathrm{eV}$

From Bragg's law,

$2\mathrm{dsin\theta }=\mathrm{n\lambda }$

,$$\begin{gathered} \mathrm{For}\mathrm{first}\mathrm{maxima},\mathrm{n}=1 \\ =2\mathrm{dsin\theta }=\mathrm{\lambda } \\ =2\mathrm{dsin\theta }=\mathrm{\lambda } \\ \because \frac{1}{2}{\mathrm{mv}}^2=\mathrm{E} \\ \mathrm{p}=\sqrt{2\mathrm{Em}} \\ \mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2\mathrm{mE}}} \\ 2\mathrm{Em}=\frac{{\mathrm{h}}^2}{4{\mathrm{d}}^2{\sin }^2\mathrm{\theta }} \\ \mathrm{E}=\frac{{\mathrm{h}}^2}{8{\mathrm{md}}^2{\sin }^2\mathrm{\theta }} \\ \mathrm{Given}, \\ \mathrm{d}=1\stackrel{\circ }{\mathrm{A}} \\ ={10}^{-10}\mathrm{m} \\ \mathrm{\theta }=60° \\ \mathrm{h}=6.64\times {10}^{-34}\mathrm{Js} \\ \mathrm{m}=9.1\times {10}^{-31}\mathrm{kg} \end{gathered}$$

Substituting these values in the above equation,

We get :

$$\begin{gathered} \mathrm{E}=\frac{{\left(6.64\times {10}^{-34}\right)}^2}{8\times 9.1\times \left({10}^{-10}\right){\sin }^{2}60°} \\ =\frac{44.0896\times {10}^{-68}}{8\times 9.1\times {10}^{-51}\times \left({\displaystyle \frac{3}{4}}\right)}=0.8075\times {10}^{-17}\mathrm{J} \\ =\frac{0.8075\times {10}^{-17}}{1.6\times {10}^{-19}}\mathrm{eV} \\ =\mathbf{50}\mathbf{.}\mathbf{47}\mathbf{eV}\mathbf{\approx }\mathbf{50}\mathbf{eV} \end{gathered}$$