Silver crystallizes in fcc lattice . If edge length of the unit cell is 4.077x10-8cm then calculate radius of atom

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Solution

Dear student ,

For the fcc unit cell R = $\frac{a}{2 \sqrt{2}}$ = 0.3535 a = 0.3535 x 4.077 x 10^-8 = 1,44 x 10^-8 cm
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Silver crystallizes in fcc lattice . If edge length of the unit cell is 4.077x10-8cm then calculate radius of atom

Dear student , For the fcc unit cell R = a22 = 0.3535 a = 0.3535 x 4.077 x 10-8 = 1,44 x 10-8 cm Regards

Dear student ,

For the fcc unit cell R = $\frac{a}{2 \sqrt{2}}$ = 0.3535 a = 0.3535 x 4.077 x 10^-8 = 1,44 x 10^-8 cm
Regards