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(B) The condition for constructive interference in crystal diffraction (Bragg's Law) is given by $2d \sin 	heta = n \lambda$, where $	heta$ is the g

Vedclass · Accepted Answer

$50$

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(B) The condition for constructive interference in crystal diffraction (Bragg's Law) is given by $2d \sin 	heta = n \lambda$, where $	heta$ is the glancing angle.From the figure, the angle of incidence $i$ is measured with respect to the normal. The glancing angle $	heta$ is the angle between the incident ray and the crystal plane, so $	heta = 90^{\circ} - i$.Given $i = 30^{\circ}$, we have $	heta = 90^{\circ} - 30^{\circ} = 60^{\circ}$.For first-order diffraction, $n = 1$. Substituting the values:$2 	imes (1 	imes 10^{-10}\; m) 	imes \sin(60^{\circ}) = 1 	imes \lambda$$\lambda = 2 	imes 10^{-10} 	imes \frac{\sqrt{3}}{2} = \sqrt{3} 	imes 10^{-10}\; m \approx 1.732 	imes 10^{-10}\; m$.The de-Broglie wavelength of an electron accelerated by potential $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}}$.Squaring both sides:$\lambda^2 = \frac{h^2}{2meV} \Rightarrow V = \frac{h^2}{2me\lambda^2}$.Substituting the values:$V = \frac{(6.6 	imes 10^{-34})^2}{2 	imes 9.1 	imes 10^{-31} 	imes 1.6 	imes 10^{-19} 	imes (\sqrt{3} 	imes 10^{-10})^2}$$V = \frac{43.56 	imes 10^{-68}}{29.12 	imes 10^{-50} 	imes 3} = \frac{43.56 	imes 10^{-68}}{87.36 	imes 10^{-50}} \approx 0.4986 	imes 10^{18} 	imes 10^{-68} 	imes 10^{50} \approx 49.86\; V$.Thus, $V$ should be about $50\; V$.

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