Question

The no.of unit cell in 1m3 of fcc nickel is (radius of nickel is 1.243 angstrom )

Answer 1

Answer:2.3×10^28

Explanation:

Answer 2

Answer:

The no.of unit cell in 1m3 of fcc nickel is 1.22×10³².

Explanation:

Given : Radius of nikel = 1.243 Angstrom

            volume of nikle = 1m³

Now for Fcc nikle the realtion between the radius and lattice constant is

     $4r=$ $\sqrt{2} a$

r= radius of the atom

a= latiice constant

so we cal calculate lattice constant from above

$4r= \sqrt{2} a\\\\a=\frac{4r}{\sqrt{2} } \\\\a= \frac{4\times 1.243}{\sqrt{2} }\\\\ a=\frac{4.972}{1.141}\\\\a=4.357 \ \ Angstrom$

now volume of lattice cell = a³

volume = (4.35)³ × (10⁻³⁰)

Now number of unit cell = $\frac{volume \ \of \ \sample }{volume\ \of\ \ each\ \ unit \ \ cell}$

 $N= \frac{1}{81.74\times 10^{-30} } \ \\\\N =1.22 \times 10^{32}$

Hence the Number of unit cell in 1m³ sample is 1.22×10³².