Question:
A beam of electrons of energy $E$ scatters from a target having atomic spacing of $1 A$. The first maximum intensity occurs at $\theta=60^{\circ}$. Then $\mathrm{E}$ (in $\mathrm{eV}$ ) is
(Planck constant $\mathrm{h}=6.64 \times 10^{-34} \mathrm{Js}$, $1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}$, electron mass $\mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}$ )
Solution:
$2 \mathrm{~d} \sin \theta=\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$
$2 \times 10^{-10} \times \frac{\sqrt{3}}{2}=\frac{6.6 \times 10^{-34}}{\sqrt{2 \mathrm{mE}}}$
$\mathrm{E}=\frac{1}{2} \times \frac{6.64^{2} \times 10^{-48}}{9.1 \times 10^{-31} \times 3 \times 1.6 \times 10^{-19}}=50.47$