Question

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

Answer 1

From Bragg's equation $2d\sin \theta =\lambda$ and de-Broglie wave length, $\lambda =\frac{h}{P}=\frac{h}{\sqrt{2mE}}$


$2d\sin \theta =\lambda =\frac{h}{\sqrt{2mE}}$

$\Rightarrow 2\times {10}^{-10}\times \frac{\sqrt{3}}{2}=\frac{6.6\times {10}^{-34}}{\sqrt{2mE}}$

$[\because \theta =60°$ and $d=1\stackrel{\circ }{\text{A}}=1\times {10}^{-10}\text{m}]$

$\therefore E=\frac{1}{2}\times \frac{{6.64}^2\times {10}^{-48}}{9.1\times {10}^{-31}\times 3\times 1.6\times {10}^{-19}}\simeq 50\text{eV}$