Question

1. lead adopts the face - centered cubit unit cell in its crystal structure. if the edge length of the unit cell is 495 pm, find the atomic radius of lead.

Explanation:

Step1: Recall the relationship for FCC unit cell

For a face - centered cubic (FCC) unit cell, the relationship between the edge length \(a\) and the atomic radius \(r\) is \(a = 2\sqrt{2}r\).

Step2: Solve for \(r\)

We can re - arrange the formula \(a = 2\sqrt{2}r\) to solve for \(r\). So, \(r=\frac{a}{2\sqrt{2}}\).
Given \(a = 495\space pm\), substitute \(a\) into the formula:

$$
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$$r\approx175\space pm$$

Answer:

The atomic radius of Lead is approximately \(175\space pm\)