The first order reflection $(n = 1)$ from a crystal of the X-ray from a copper anode tube $\left( {\lambda = 1.54\mathop {\rm A}\limits^ \circ \;} \right)$ occurs at an angle of $45^\circ $. What is the distance between the set of plane causing the diffraction?

$0.905\;\mathop {\rm A}\limits^ \circ \;$

$1.089 \times {10^{ - 9}}\;{\rm{m}}$

JEE CHEMISTRY MCQ: SOLID STATE

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The first order reflection

(n = 1)

from a crystal of the X-ray from a copper anode tube

\left( {\lambda = 1.54\mathop \mathrm{A}\limits^ \circ \;} \right)

occurs at an angle of

45^\circ

. What is the distance between the set of plane causing the diffraction?

View answer and explanation

Correct option: C

0.905\;\mathop \mathrm{A}\limits^ \circ \;

The correct answer is C (

0.905\;\mathop \mathrm{A}\limits^ \circ \;

) because it is calculated using Bragg's Law,

n\lambda = 2d\sin{\theta}

, with the given wavelength and angle.