Silicon has the diamond structure with a unit cell edge length of 5.43 Å and eight atoms per unit cell. (b) Suppose you dope that 1 cm3 sample of silicon with 1 ppm of phosphorus that will increase the conductivity by a factor of a million. How many milligrams of phosphorus are required?

Video transcript

Hello everyone. So in this video we're calculating for the mass of boron in milligrams. If a five centimeter cube sample of the semiconductor is at two PPm. So first we need to go and calculate for the volume of Yoon itself That's going to equal to 5.65 Angstrom cute Putting this into my calculator. The numerical value is 180.36 by unit being Angstrom cute. Now let's go ahead and convert the sample convert volume of sample. So we have that five centimeter cubed sample. We're gonna want to convert that into angstrom. So the direct conversion is for every one centimeter will have 10 to the eight angstrom. And we're gonna go ahead and and cube all of this because we want the units of centimeter cubes to match. So once we do that we'll get the final value of five times 10-24 power angstrom cute. Now going ahead to calculate for our G. E. So gallium Adams We'll have five times 10 to the 24 Angstrom. So for every eight items because we have the geranium cubic cell unit at eight atoms per unit cell 28 Atoms 4 1 80.36. Ancient cute. So you can see here that the units of extreme cute will cancel leaving us with the numerical value of 2.218 times 10-23rd power of G. E. Atoms. Now let's recall the information that was given. We have this right here that just means we have one b Per 10 to the 6G of course this is going to be a very important conversion factor. So just scrolling down just a little bit, we'll have Again using this value to start our conversion of 2. times 10 to the 23rd G. E. Adams. And then using that conversion factor now every to be adam we'll have 10 to the six G. E atoms. So you can see that unit will counsel numerical value will be 4.436 times 10 - 17 units being be atoms. Now to finally calculate for our mass. Let's keep in mind the molar mass of B Which can be found in a periodic table, that's gonna be 10.81 g per mole of course, again that would be a converted factor as well. So starting off with our atoms that we just saw for up top here. So we have 4.43, 6 times 10 to the 17 power of bi atoms. We're gonna go ahead and use avocados number to go ahead and convert our atoms into more. So for every one mole that we have, we of course have 6.22 times 10 to the 23rd atoms. Alright now going ahead to use our molar mass, we have 10.81 g on top and one mole on the bottom. The reason for this is we're trying to get rid of the mole units to go into grams. We want the final answer to be in milligrams. So we'll just convert our grams to milligrams. So for every one mg that we have, we have 10 to the negative third grams. Let's go ahead and double checked. So we have the B atoms canceling our moles and grams. So then maybe put all this into calculator. You'll get 7.96 times 10 to the negative three mg. And that's going to be my final answer for this question. Thank you all so much for watching.