When a certain crystal was studied by the Bragg technique using X-rays of wavelength 229 pm, an X-ray reflection was observed at an angle of $23{^\circ}$ 20’. What is the corresponding interplanar spacing? $\left\lbrack \sin{\left( 23{^\circ}20^{'} \right) = 0.396} \right\rbrack$

Question

Dr. Abubacker · Accepted Answer

(c)
Given that,
$\lambda = 229\ pm\$and$\ 	heta = 23{^\circ}20'$
Substituting these values in the Bragg’s equation, we have
$d_{hkl} = \frac{\lambda}{2\sin	heta} = \frac{229\ pm}{2\sin{(23{^\circ}20')}}$
$= \frac{229\ pm}{2 	imes \ 0.396}$
= 289.2 pm

Q. When a certain crystal was studied by the Bragg technique using X-rays of wavelength 229 pm, an X-ray reflection was observed at an angle of $23{^\circ}$ 20’. What is the corresponding interplanar spacing? $\left\lbrack \sin{\left( 23{^\circ}20^{'} \right) = 0.396} \right\rbrack$

Explanation

Reviewed by Dr. Abubacker

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