The distance between interatomic lattice plan | vedclass

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(B) Given the distance between lattice planes $d = 10\ \mathring A$. According to Bragg's Law,$2d \sin 	heta = n \lambda$. To find the maximum wavele

Vedclass · Accepted Answer

$20$

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(B) Given the distance between lattice planes $d = 10\ \mathring A$. According to Bragg's Law,$2d \sin 	heta = n \lambda$. To find the maximum wavelength $\lambda_{\max}$,we set the order of diffraction $n = 1$ and the maximum value of $\sin 	heta = 1$. Substituting these values: $\lambda_{\max} = \frac{2d \sin 	heta}{n} = \frac{2 	imes 10\ \mathring A 	imes 1}{1} = 20\ \mathring A$.

The distance between interatomic lattice planes is $10\ \mathring A$. The maximum wavelength of $X$-rays that can be diffracted by this crystal is ......$\mathring A$.

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