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(C) According to Bragg's law for diffraction, $2d \sin 	heta = n\lambda$. For the first maximum, $n = 1$, so $2d \sin 	heta = \lambda$.Given $d = 1

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(C) According to Bragg's law for diffraction, $2d \sin 	heta = n\lambda$. For the first maximum, $n = 1$, so $2d \sin 	heta = \lambda$.Given $d = 1 \, Å = 10^{-10} \, m$ and $	heta = 60^{\circ}$.Thus, $\lambda = 2 	imes 10^{-10} 	imes \sin(60^{\circ}) = 2 	imes 10^{-10} 	imes \frac{\sqrt{3}}{2} = \sqrt{3} 	imes 10^{-10} \, m$.Using the de Broglie wavelength formula, $\lambda = \frac{h}{\sqrt{2mE}}$, we have $\sqrt{2mE} = \frac{h}{\lambda}$.Squaring both sides, $2mE = \frac{h^2}{\lambda^2}$, so $E = \frac{h^2}{2m\lambda^2}$.Substituting the values: $E = \frac{(6.64 	imes 10^{-34})^2}{2 	imes 9.1 	imes 10^{-31} 	imes (\sqrt{3} 	imes 10^{-10})^2} = \frac{44.0896 	imes 10^{-68}}{2 	imes 9.1 	imes 10^{-31} 	imes 3 	imes 10^{-20}} = \frac{44.0896 	imes 10^{-68}}{54.6 	imes 10^{-51}} \approx 0.8075 	imes 10^{-17} \, J$.To convert to $eV$, divide by $1.6 	imes 10^{-19} \, J/eV$:$E = \frac{0.8075 	imes 10^{-17}}{1.6 	imes 10^{-19}} \approx 50.47 \, eV$.Rounding to the nearest integer, $E \approx 50 \, eV$.

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