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A reflection from (111) planes of a cubic crystal was observed at a glancing angle of 11.2^@ when X-rays of wavelength 154 pm were used. What is the length of the side of the unit cell ? (sin 11.2^@=0.1944) |
Answer» <html><body><p></p>Solution :Here, `lambda`=154 <a href="https://interviewquestions.tuteehub.com/tag/pm-1156895" style="font-weight:bold;" target="_blank" title="Click to know more about PM">PM</a>, n=1 , `<a href="https://interviewquestions.tuteehub.com/tag/theta-1412757" style="font-weight:bold;" target="_blank" title="Click to know more about THETA">THETA</a>=11.2^@` <br/> Applying Bragg's equation `2 d <a href="https://interviewquestions.tuteehub.com/tag/sin-1208945" style="font-weight:bold;" target="_blank" title="Click to know more about SIN">SIN</a> theta= n lambda` <br/> i.e., `d_111=(nlambda)/(2 sin theta)=(1xx154)/(2xx0.1944) ` pm =396 pm <br/> Further , the separation between the <a href="https://interviewquestions.tuteehub.com/tag/planes-1155607" style="font-weight:bold;" target="_blank" title="Click to know more about PLANES">PLANES</a> of a <a href="https://interviewquestions.tuteehub.com/tag/cubic-940181" style="font-weight:bold;" target="_blank" title="Click to know more about CUBIC">CUBIC</a> crystal is given by <br/> `d_"hkl"=a/(sqrt(h^2+k^2+l^2))` <br/> `therefore d_11=a/(sqrt(1^2+1^2+1^2))`=396 pm (calculated above ) <br/> or `a=396xxsqrt3`=686 pm</body></html> | |