Substitutional impurity in a lattice?

[orangeincup]

Homework Statement

Boron atoms at a concentration of 3.60E+16 atoms/cm3 are added to pure intrinsic silicon as a substitutional impurity. Assume that the boron atoms are distributed homogeneously through the silicon in a cubic array.

Enter the fraction, by weight, accounted for by impurity atoms in the lattice. To do this, you will need to remember that silicon has a diamond structure with a lattice constant of 5.43 angstroms

Homework Equations

8/(a)^3=volume density

The Attempt at a Solution

8/(a)^3=4.99*10^23

3.60*10^22+4.99*10^23/4.99*10^23 = .99

 

[phyzguy]

I have no idea what you have attempted to do in your calculation. Some words would help. How many silicon atoms are in a cm^3, and how did you calculate it? How many boron atoms are in the same cm^3? What is the total weight of these silicon atoms? What is the total weight of the boron atoms? The ratio of the two will give you the fraction by weight.

 

[Funestis]

Homework Statement

Boron atoms at a concentration of 3.60E+16 atoms/cm3 are added to pure intrinsic silicon as a substitutional impurity. Assume that the boron atoms are distributed homogeneously through the silicon in a cubic array.

Enter the fraction, by weight, accounted for by impurity atoms in the lattice. To do this, you will need to remember that silicon has a diamond structure with a lattice constant of 5.43 angstroms

Homework Equations

8/(a)^3=volume density

The Attempt at a Solution

8/(a)^3=4.99*10^23

3.60*10^22+4.99*10^23/4.99*10^23 = .99

I have no idea what you have attempted to do in your calculation. Some words would help. How many silicon atoms are in a cm^3, and how did you calculate it? How many boron atoms are in the same cm^3? What is the total weight of these silicon atoms? What is the total weight of the boron atoms? The ratio of the two will give you the fraction by weight.

Since the weights of silicon and boron atoms are essentially a given, I think the hard part of your problem is finding how many silicon atoms there are in a cubic centimeter of the material. If I remember rightly there are eight atoms in the unit cell for a diamond cubic crystal, that combined with the lattice constant should help to get you there.