A beam of electrons of energy E scatters from a target having atomic spacing of $1 \AA$. The first maximum intensity occurs at $\theta=60^{\circ}$. Then $\mathrm{E}$ (in $\mathrm{eV}$ ) is (Plank constant $h=6.64 \times 10^{-34} \mathrm{Js}, 1 \overline{\mathrm{eV}=1} .6 \times 10^{-19} \mathrm{~J}$, electron mass $m=9.1 \times 10^{-31} \mathrm{~kg}$ )
(50)
From Bragg's equation $2 d \sin \theta=\lambda$ and de-Broglie
wavelength, $\lambda=\frac{h}{P}=\frac{h}{\sqrt{2 m E}}$
'
$2 d \sin \theta=\lambda=\frac{h}{\sqrt{2 m E}}$
$\Rightarrow 2 \times 10^{-10} \times \frac{\sqrt{3}}{2}=\frac{6.6 \times 10^{-34}}{\sqrt{2 m E}}$
$\left[\because \theta=60^{\circ}\right.$ and $\left.d=1 A=1 \times 10^{-10} \mathrm{~m}\right]$
$\therefore E=\frac{1}{2} \times \frac{6.64^{2} \times 10^{-48}}{9.1 \times 10^{-31} \times 3 \times 1.6 \times 10^{-19}} \simeq 50 \mathrm{eV}$
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