A beam of electrons of energy E scatters from a target having atomic spacing of 1 ∖ AAA. The first maximum intensity occurs at =60^ . Then E (in eV ) is (Planck constant h=6.64 10^-34 Js ,1 eV=1.6 10…

Question

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

Accepted Answer

Correct answer: 50.47. (50.47) JEE Main 2020 12 mv^2=E ⇒P= 2 Em ∴ = h 2 Em For first maxima, 2 dsin ⁡ = ⇒2 10^-10 32= h 2 Em ⇒2 Em= (10^10 h )^23 ⇒E= (10^10 h )^26m_e e ( eV ) ⇒E=50.47eV

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

Solution

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