# 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 $E$ (in eV) is _______. (Plank constant $h=6.64\times 10^{-34}Js,$                            $1\; eV=1.6\times 10^{-19}J,$ electron mass $m=9.1\times 10^{-31}kg$) Option: 1 50.47 Option: 2 100.94 Option: 3 40.47 Option: 4 80

$\begin{array}{l} 2 \mathrm{~d} \sin \theta=\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}} \\ \\ 2 \times 10^{-10} \times \frac{\sqrt{3}}{2}=\frac{6.6 \times 10^{-34}}{\sqrt{2 \mathrm{mE}}} \\ \\ \mathrm{E}=\frac{1}{2} \times \frac{6.64^{2} \times 10^{-48}}{9.1 \times 10^{-31} \times 3 \times 1.6 \times 10^{-19}}=50.47 \end{array}$

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