Q72 / 75 · JEE Main 2020 (05 Sep Shift 1) · Physics

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

Correct answer: 50

Solution

$2 dsin θ = n λ$

$2 d \frac{\sqrt{3}}{2} = (1) λ$

$d = 1 Å$

$λ = \sqrt{3} Å$

$λ = \sqrt{3} = \sqrt{\frac{150}{V}}$

$V = 50 volt$

$E = 50 eV$

Hence, the correct answer is 50.