If a face-centered cubic unit cell has a volume of 1.45 $\times$ 10⁸ pm³ and the atoms at the corners touch the atom on the face,what must be the atom's radius?

Question

Quizplus · Accepted Answer

First, we need to find the length of the unit cell edge (a). Since each atom on the corner contributes 1/8 to the unit cell, and each atom on the face contributes 1/2 to the unit cell, we have:

4 x (1/8) + 1 x (1/2) = 1

Solving for a, we get:

a = (4/3)^1/2 x r

where r is the radius of the atom.

Next, we can find the volume of the unit cell:

V = a^3

Plugging in the given value of V, we get:

1.45 x 10^-22 cm^3 = [(4/3)^1/2 x r]^3

Solving for r, we get:

r = (3V/4π)^1/3

r = (3 x 1.45 x 10^-22 cm^3/4π)^1/3

r ≈ 186 pm.

Therefore, the correct answer is A.

If a Face-Centered Cubic Unit Cell Has a Volume of 1.45