A beam of electrons of energy E scatters from a target having atomic spacing of 1 \Å. The first maximum intensity occurs at \\theta=60^\circ . Then {E} (in eV) is _______. (Planck constant {h}=6.64\times10^{-34} {Js},1 {eV}=1.6 \times 10^{-19} { J}, electron mass {m}=9.1 \times 10^{-31} { kg} ) [5 Sep 2020 Shift 1]

JEE Mains 05-Sep-2020 Shift 1 Solved Paper

Section: Physics

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

[5 Sep 2020 Shift 1]

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