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  4. A beam of electrons of energy E scatters from a target having atomic spacing of 1 mathringA The first maximum intensity occurs at θ=60° Then E (in eV ) is . (Planck constant h =6.64 × 10-34 Js 1 eV =1.6 × 10-19 J , electron mass . m =9.1 × 10-31 kg )

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Q. A beam of electrons of energy E scatters from a target having atomic spacing of $1 \,\mathring{A}$ The first maximum intensity occurs at $\theta=60^{\circ}$ Then E (in eV ) is ____. (Planck constant $h =6.64 \times 10^{-34} \,Js$ $1 eV =1.6 \times 10^{-19} J $, electron mass $\left. m =9.1 \times 10^{-31} \,kg \right)$

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Solution:

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$2 d \sin \theta=\lambda=\frac{ h }{\sqrt{2 mE }}$
$2 \times 10^{-10} \times \frac{\sqrt{3}}{2}=\frac{6.6 \times 10^{-34}}{\sqrt{2 mE }}$
$E=\frac{1}{2} \times \frac{6.64^{2} \times 10^{-43}}{9.1 \times 10^{-31} \times 3 \times 1.6 \times 10^{-19}}=50.47$