An X-ray beam of wavelength 1 mathring A is | vedclass

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(D) According to Bragg's law,the condition for diffraction is given by $2d \sin 	heta = n\lambda$. Given: Lattice spacing $d = 2.8 \ \mathring A = 2.

Vedclass · Accepted Answer

$10.3$

Vedclass · Answer

(D) According to Bragg's law,the condition for diffraction is given by $2d \sin 	heta = n\lambda$. Given: Lattice spacing $d = 2.8 \ \mathring A = 2.8 	imes 10^{-10} \ m$. Wavelength $\lambda = 1 \ \mathring A = 1 	imes 10^{-10} \ m$. Order of diffraction $n = 1$. Substituting the values into the formula: $2 	imes (2.8 	imes 10^{-10} \ m) 	imes \sin 	heta = 1 	imes (1 	imes 10^{-10} \ m)$. $5.6 	imes 10^{-10} 	imes \sin 	heta = 1 	imes 10^{-10}$. $\sin 	heta = \frac{1}{5.6} \approx 0.1786$. $	heta = \sin^{-1}(0.1786) \approx 10.27^\circ$. Rounding to the nearest provided option,we get $	heta \approx 10.3^\circ$.

An $X$-ray beam of wavelength $1 \ \mathring A$ is incident on a crystal with a lattice spacing of $2.8 \ \mathring A$. Find the value of the Bragg angle in degrees for the first-order diffraction.

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