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  • A beam of electrons of energy scatters from a target having atomic spacing of . The first maximum inte

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

Answers (1)

best_answer

\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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Deependra Verma

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