Silver crystallizes with f.c.c. unit cell. Each side of the unit cell has a length of 409 pm. What is the radius of Ag atom? ?

To determine the radius of a silver (Ag) atom when it crystallizes in a face-centered cubic (f.c.c.) structure, we can use the relationship between the unit cell dimensions and the atomic radius. In an f.c.c. lattice, atoms are located at each corner of the cube and at the centers of each face. This arrangement leads to a specific geometric relationship that we can leverage.

Understanding the f.c.c. Structure

In a face-centered cubic unit cell, the atoms touch each other along the face diagonal. The face diagonal can be expressed in terms of the unit cell edge length (a) and the atomic radius (r). The formula for the face diagonal (d) in terms of the edge length is:

• d = √2 * a

Since the face diagonal also equals four times the atomic radius (because there are two atomic radii from each of the two atoms at the ends of the diagonal), we can set up the equation:

• d = 4r

Calculating the Atomic Radius

Now, we can combine these two relationships. First, we substitute the expression for the face diagonal into the equation:

• √2 * a = 4r

Next, we can solve for the atomic radius (r):

• r = (√2 * a) / 4

Given that the edge length (a) of the silver unit cell is 409 pm (picometers), we can plug this value into our equation:

• r = (√2 * 409 pm) / 4

Performing the Calculation

Now, let's calculate the radius step by step:

• First, calculate √2, which is approximately 1.414.
• Next, multiply this by 409 pm:
• 1.414 * 409 pm ≈ 578.5 pm.
• Now, divide this result by 4:
• r ≈ 578.5 pm / 4 ≈ 144.625 pm.

Final Result

Thus, the radius of a silver atom in a face-centered cubic structure is approximately 144.6 pm. This value is consistent with the known atomic radius of silver, which is around 144 pm, confirming the accuracy of our calculations.