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Two lines A and B, of an x-ray beam give first order reflection maximum at a glancing angle of 30^{\circ} and third order reflection maximum at an angle of 60^{\circ}(\pi / 3) respectively from the face of the same crystal. If the wave length of line B is 0.098\mathrm{hm}, the the wave length of line A is -

Option: 1

0.092 \mathrm{~nm}


Option: 2

0.056 \mathrm{~nm}


Option: 3

0.126 \mathrm{~nm}


Option: 4

0.170 \mathrm{~nm}


Answers (1)

best_answer

 From Bragg's diffraction -

\mathrm{ 2 d \sin \theta=n \lambda}
\mathrm{ Here, \theta_1=30^{\circ}, \lambda_1= ?,n_1=1 \theta_2=60^{\circ}, \lambda_2=0.098 \mathrm{1m}, n_1=3 }
\mathrm{Now, \begin{aligned} \frac{n_1 \lambda_1}{\sin \theta_1} & =\frac{n_2 \lambda_2}{\sin \theta_2} \\ \frac{1 \times \lambda_1}{\sin 30^{\circ}} & =\frac{3 \times 0.098}{\sin 60^{\circ}} \\ \lambda_1=\frac{3 \times 0.098}{\sqrt{3} / 2} & =\frac{6 \times 0.098 \times \sqrt{3}}{3} \\ & =2 \times 0.098 \times \sqrt{3} \\ \lambda_1 & =2 \times 0.098 \times 1.732 \\ \lambda_1 & =0.1702 \mathrm{~nm} \end{aligned} }

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Kuldeep Maurya

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