Cesium crystallises in a body-centred cubic lattice with cell edge, a= 628 pm. Calculate the radius of Cs atom.

272 pm

In a body-centered cubic (BCC) lattice, the relationship between the edge length of the unit cell (\(a\)) and the atomic radius (\(r\)) can be given by the formula:

\(a = 4 \sqrt{2} r\)

Given that the cell edge length (\(a\)) is 628 pm, we can rearrange the formula to solve for the radius (\(r\)):

\(r = \frac{a}{4 \sqrt{2}} = \frac{628 \, \text{pm}}{4 \sqrt{2}} \approx 157 \, \text{pm}\)

So, the radius of the Cs atom is approximately 157 pm. Therefore, the correct answer is: 1. 157 pm