Cesium crystallises in a body-centred cubic lattice with cell edge, a= 628 pm. Calculate the radius of Cs atom.
Answer & explanation
Correct answer: option 1
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