An element having an atomic radius of 0.14 nm crystallizes in an $$f_{cc}$$ unit cell. What is the length of a side of the cell?

An element with an atomic radius of 0.14 nm crystallizes in a face-centered cubic (fcc) unit cell. In an fcc lattice, atoms touch along the face diagonal of the cube. The face diagonal is composed of two atomic diameters because it passes through two atoms: one at a corner and one at the center of the face. Therefore, the length of the face diagonal is equal to four times the atomic radius.

Given the atomic radius $$ r = 0.14 $$ nm, the face diagonal is calculated as follows:

Face diagonal $$ = 4 \times r = 4 \times 0.14 = 0.56 $$ nm.

For a cubic unit cell with side length $$ a $$, the face diagonal can also be found using the Pythagorean theorem. The face is a square with side $$ a $$, so the diagonal is $$ \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2} $$.

Setting this equal to the face diagonal from the atomic arrangement:

$$ a\sqrt{2} = 0.56 $$ nm.

To solve for $$ a $$, divide both sides by $$ \sqrt{2} $$:

$$ a = \frac{0.56}{\sqrt{2}} $$ nm.

Rationalize the denominator by multiplying the numerator and denominator by $$ \sqrt{2} $$:

$$ a = \frac{0.56 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{0.56 \times \sqrt{2}}{2} = 0.28 \times \sqrt{2} $$ nm.

Now, substitute the approximate value of $$ \sqrt{2} \approx 1.4142 $$:

$$ a = 0.28 \times 1.4142 = 0.395976 $$ nm.

Rounding 0.395976 nm to two decimal places gives 0.40 nm, as the digit in the thousandths place is 5, which rounds up the hundredths place from 9 to 10, resulting in 0.40 nm.

Comparing with the options:

A. 0.56 nm (which is the face diagonal, not the side)

B. 0.24 nm (incorrect)

C. 0.96 nm (incorrect)

D. 0.4 nm (matches the calculated side length)

Hence, the correct answer is Option D.