# Caesium has an atomic radius of #2.5 xx10^(-8)# cm and crystallizes in a body-centered cubic crystal structure . The length of the unit cell edge in picometers is?

##### 1 Answer

#### Explanation:

All you have to do here is use the configuration of the unit cell for a **body-centered cubic** crustal structure and the radius of a cesium atom to help you find the *diagonal* of the unit cell.

Once you ahve the diagonal of the cell, you can use the Pythagorean Theorem to determine the length of the unit cell.

So, a **body-centered cubic** crystal structure is characterized by a total of *lattice point*

one lattice pointin every corner of the cubeone lattice pointin the center of the cube

To find the diagonal of the cell, pick a **face** of the cube and draw it as a *square*. You should end up with something like this

Now, let's say that

Using Pythagoras' Theorem, you can say that

#d^2 = x^2 + x^2#

#d^2 = 2x^2" " " "color(purple)((1))#

Now focus on writing the diagonal of the square by using *radius* of a cesium atom. Notice that you have, starting from the top-right of the square

#r# ,theradiusof the top-right cesium atom#2r# ,the diameterof the center atom#r# ,theradiusof the low-left cesium atom

The diagonal of the square will thus be

#d = r + 2r + r = 4r#

Plug this into equation

#(4r)^2 = 2x^2#

#16r^2 = 2x^2#

#8r^2 = x^2 implies x = sqrt(8r^2) = r * 2sqrt(2)#

You can find the length of the unit cell in *centimeters*, then convert it to *picometers*

#x = 2.5 * 10^(-8)"cm" * 2sqrt(2) = 7.1 * 10^(-8)"cm"#

As you know, you can use the following conversion factor

#"1 cm" = 10^(-2)"m" = 10^(-2) xx 10^(12)"pm" = 10^(10)"pm"#

This means that the edge length of the unit cell will be

#7.1 * 10^(-8)color(red)(cancel(color(black)("cm"))) * (10^(10)"pm")/(1color(red)(cancel(color(black)("cm")))) = color(green)("710 pm")#