Copper crystallises in fcc with a unit length of 361 pm. What is the radius of copper atom?

Copper crystallizes in a face-centered cubic (fcc) lattice with a unit cell edge length of 361 pm. We need to find the radius of the copper atom.

In an fcc lattice, atoms are located at each corner of the cube and at the center of each face. The atoms touch along the face diagonal of the cube. This means the face diagonal is equal to four times the atomic radius (r).

The face diagonal of a cube can be found using the Pythagorean theorem. For a cube with edge length a, the face diagonal is the diagonal of a square face with side a. The length of the diagonal is given by $$\sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2}$$.

Since the face diagonal is occupied by atoms touching each other, it is equal to 4r. Therefore, we set up the equation:

$$a\sqrt{2} = 4r$$

We are given that the edge length a is 361 pm. Substitute this value into the equation:

$$361 \times \sqrt{2} = 4r$$

Now, solve for r. First, isolate r by dividing both sides by 4:

$$r = \frac{361 \times \sqrt{2}}{4}$$

We know that $$\sqrt{2} \approx 1.4142$$. Substitute this value:

$$r = \frac{361 \times 1.4142}{4}$$

First, multiply 361 by 1.4142:

$$361 \times 1.4142 = 361 \times (1 + 0.4 + 0.014 + 0.0002) = 361 \times 1 + 361 \times 0.4 + 361 \times 0.014 + 361 \times 0.0002$$

Calculate each part:

$$361 \times 1 = 361$$

$$361 \times 0.4 = 144.4$$

$$361 \times 0.014 = 361 \times \frac{14}{1000} = \frac{5054}{1000} = 5.054$$

$$361 \times 0.0002 = 0.0722$$

Add these together:

$$361 + 144.4 = 505.4$$

$$505.4 + 5.054 = 510.454$$

$$510.454 + 0.0722 = 510.5262$$

So,

$$361 \times 1.4142 = 510.5262$$

Now divide by 4:

$$r = \frac{510.5262}{4} = 127.63155 \text{ pm}$$

Rounding to the nearest whole number, we get approximately 128 pm.

Therefore, the radius of the copper atom is 128 pm.

Looking at the options:

A. 157 pm

B. 128 pm

C. 108 pm

D. 181 pm

Hence, the correct answer is Option B.