JEE PYQ: Wave Optics Question 15
Question 15 - 2020 (05 Sep Shift 1)
A beam of electrons of energy E scatters from a target having atomic spacing of 1$\AA$. The first maximum intensity occurs at $\theta = 60°$. 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)
Show Answer
Answer: 50
Solution
From Bragg’s equation $2d\sin\theta = \lambda$ and de-Broglie wavelength, $\lambda = \frac{h}{P} = \frac{h}{\sqrt{2mE}}$
$2d\sin\theta = \frac{h}{\sqrt{2mE}}$
$\Rightarrow 2 \times 10^{-10} \times \frac{\sqrt{3}}{2} = \frac{6.6 \times 10^{-34}}{\sqrt{2mE}}$
$[\because \theta = 60°$ and $d = 1\AA = 1 \times 10^{-10}$ m]
$\therefore E = \frac{1}{2} \times \frac{6.64^2 \times 10^{-68}}{9.1 \times 10^{-31} \times 3 \times 1.6 \times 10^{-19}} = 50$ eV
BETA
