Question

Copper crystallizes in FCC with a unit cell length of 361pm. What is the radius (in pm) of copper atoms?
(A) 109
(B) 127
(C) 157
(D) 181

Answer 1

Hint: The atoms are arranged at the corners of the cube and at the centers of the faces in the FCC lattice. We can also say that the face diagonal of the unit cell is of length $\sqrt 2 a$ where a is the unit cell length.

Complete step by step solution:
We are given that copper crystallizes in FCC lattice arrangement.
- As copper is crystallized in FCC, we can say that copper atoms will be there at the corners of the cube as well as at the faces of the cube.
- Now, we are given the length of the unit cell.
- We can say that the face diagonal of the cube will be $\sqrt 2 a$. We obtained this using Pythagoras theorem as ${\text{face diagonal = }}\sqrt {{a^2} + {a^2}} = \sqrt {2{a^2}} = \sqrt 2 a$
- Now, we obtained that value of the face diagonal of the unit cell which is $\sqrt 2 a$. Here, a is the length of the unit cell.
- Now, as we see the face of the FCC lattice, we can observe as we see the face diagonal that one whole copper atom and two half atoms of copper is involved in that face diagonal. That means the length of the face diagonal is r + r + 2r. where r is the radius of the copper atom.
So, we can write that r + r + 2r = $\sqrt 2 a$
Now, we know that a= 361 pm. So, putting that in above equation, we get
\[r + r + 2r = \sqrt 2 (361)\]
So, we can write that
\[4r = 1.41 \times 361\]
So,
\[r = \dfrac{{1.41 \times 361}}{4} = 127pm\]
Thus, we obtained that the radius of copper atoms is 127 pm.

So, the correct answer is (B).

Note: Do not get confused with the BCC lattice. BCC stands for body centered cubic cell in which the atoms are at the corners as well as in the center of the cell. FCC stands for face centered cubic lattice.