Aluminium metal forms a cubic close - packed crystal astucture. Its atomic radius is125 xx 10^(-12)m. (a) Calculatethe length of the side of the unit cell. (b) hwo many such unit cells are there in 1.00 m^(3)of aluminium ?

Aluminium metal forms a cubic close - packed crystal astucture. Its at

1.	Aluminium metal forms a cubic close - packed crystal astucture. Its atomic radius is125 xx 10^(-12)m. (a) Calculatethe length of the side of the unit cell. (b) hwo many such unit cells are there in 1.00 m^(3)of aluminium ?
Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :ccp = fcc. For fcc, `r = a/(2sqrt(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))or a = 2 sqrt2r = 2 xx 1.414 xx 125 xx 10^(-2) m = <a href="https://interviewquestions.tuteehub.com/tag/354-309041" style="font-weight:bold;" target="_blank" title="Click to know more about 354">354</a> xx10^(-12) m =354 "pm"` <br/>No. of unit cell in ` 1 m^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)= 1/(4.436 xx 10^(-29))= 2.25 xx 10^(-29) m^(3)` <br/>No. of unit cells in ` <a href="https://interviewquestions.tuteehub.com/tag/1m-283006" style="font-weight:bold;" target="_blank" title="Click to know more about 1M">1M</a> ^(3) = 1/( 4.436 xx 10^(-29)) = 2.25 xx 10^(28)`</body></html>