 Let us solve another numerical. Let's see what we have got out here. Copper atoms radius 128 picometer and mass 63.55 grams per mole, so this is some information regarding copper atoms, were found to crystallize into a cubic unit cell with an edge length of 361 picometers. If copper has a density of 8.96 grams per centimeter cube, find the kind of unit cell it forms. Interesting question. We are given some information regarding copper and we need to figure out the kind of unit cell that it forms. So let's see how I try and solve this problem. Well, if I read this question carefully, I will see that it is given that copper atoms were found to crystallize into a cubic unit cell. Now, we know that unit cells can come in different shapes. They can be cubic or orthorhombic or tetragonal etc. In fact, there are seven different shapes that are possible, but it is already given in the question that the copper atoms fall into the cubic system, right? So the unit cell that it has is definitely cubic. Now when it comes to cubic unit cells, we know that they can either be simple cubic or body centered or face centered. So the whole question boils down to figuring out if this unit cell is simple cubic or BCC or is it FCC. Now how do we figure that out? Well, in this question, we are given some information regarding the density of copper, right? So maybe I can find this out from the density. So what do we mean by density? Well, density is nothing but the mass per unit volume of a substance. So for a unit cell, the density will be the weight of the unit cell by the volume of the unit cell, right? Now what do you think will be the weight of a unit cell? For example, if I take this simple cubic unit cell which has eight atoms at these corners, what do you think will be the weight of this unit cell? Well, it won't be equal to the weight of all these atoms because these atoms are not 100% inside this unit cell, but they are in fact shared between different unit cells, right? So the weight of the unit cell will only be due to the weight of the atom that is inside the unit cell. And to figure out the number of atoms that is inside, to find out the total number of atoms that effectively lie inside a unit cell, we can always use the A effective, right? For a simple cubic unit cell, A effective will be equal to 1 by 8 into 8 because atoms at the corners are only 1 eighth inside, which comes out to be equal to 1. So we can think of a simple cubic unit cell as effectively having one atom that lies 100% inside. So the weight of the unit cell will be equal to 1 into the mass of that atom, right? Similarly, A effective of a BCC unit cell is equal to 2. So we can think of this unit cell as effectively having two atoms inside it. So its weight will be equal to 2 times the mass of the atom that make up this unit cell. Similarly, for FCC, the A effective is equal to 4. So the weight of an FCC unit cell will be 4 times the mass of the atom that makes up this unit cell. So the weight of any unit cell can be written as A effective multiplied by the mass of an atom that makes up the unit cell. And if these unit cells are cubes of edge length A, then the volume of the unit cell will be equal to A cube, right? So the density of a unit cell, the density is A effective into mass of the atom by A cube. Now if we look into this question, we will see that the density of the crystal is known, right? So we know the density, the edge length is also given, so we also know the edge length and the molar mass is also given. So the mass of an atom can also be found out. So using this information, we can find out the A effective which will be equal to the density into A cube by mass of an atom. And because it's already given in the question that the unit cell is cubic, so it can only be simple cubic or BCC or FCC, so the value of A effective can only be 1, 2 or 4, right? So on solving this, if the value comes out to be equal to 1, we can say that the unit cell is simple cubic. If it comes out to be 2, we'll call it BCC and if it's 4, then it's definitely FCC. So let us see what this comes out to be. Let me put this to the side. So now if I plug in the values, the A effective will be equal to the density which is given as 8.96 grams per centimeter cube multiplied by A cube. The edge length A is 361 picometers, whole thing cube, by the mass of an atom. Well, out here the mass is given in terms of grams per mole, so 63.55 grams is not the mass of a single copper atom but is in fact the mass of one mole of atoms. Now we require the mass of only one atom. So now if the mass of one mole which is 6.022 into 10 to the power 23, if the mass of one mole is 63.55 grams, then we need the mass of one atom, so this will be 63.55 divided by 6.022 into 10 to the power 23, this many grams, right? So the mass of an atom will be 63.55 grams divided by 6.022 into 10 to the power 23. Now if you look at this, you will realize that out here the density is measured as gram per centimeter cube but the edge length is given in terms of picometers. So let me convert these picometers in terms of centimeters. So we know that one picometer is equal to 10 to the power minus of 12 meters and this can be written as 10 to the power minus of 10 into 10 to the power minus of 2 meters and 10 to the power minus of 2 meters is equal to 1 centimeter. So one picometer is equal to 10 to the power minus of 10 centimeter, right? So if I write it out here 8.96 gram per centimeter cube into 361 into 10 to the power minus of 10 centimeter whole thing cube divided by 63.55 grams divided by 6.022 into 10 to the power 23. So let me rewrite 8.96 gram per centimeter cube into 361 into 10 to the power minus of 10 whole thing cube centimeter cube divided by 63.55 grams divided by this quantity. So I can write it as multiplied by 6.022 into 10 to the power 23, right? So this will come up here. So this gram gram and centimeter cube will cancel out. So A effective is a dimensionless quantity which will be equal to 8.96 into 361 cube into 10 to the power minus 10 cube which is 10 to the power minus of 30 into 6.022 into 10 to the power 23 divided by 63.55. Let me quickly rewrite this one more time. 10 to the power minus 30 and 10 to the power 23 is 10 to the power minus 7. So if I do the calculation out here, let me bring up my calculator. 8.96 into 361 cube into 6.022 divided by 63.55. This can be written as 3.99 into 10 to the power 1, 2, 3, 4, 5, 6, 7. So this is 3.99 into 10 to the power 7 into 10 to the power minus 7, right? So the A effective comes out to be equal to 4. So because the A effective of an FCC unit cell is 4, so the unit cell that is formed is FCC. Thank you. . .