Considering only electronegativity, rank the LED semiconductors made of solid solutions in order of increasing bandgap energy. Al0.40Ga0.60As, Al0.25Ga0.75As, Al0.05Ga0.95As

Video transcript

Hello everyone. So this problem being told to consider the electro negativity of the elements and to rank the following in extrinsic semiconductors from the least conductive to the most conductive and were given these here. So for semiconductors the difference in our electro negativity and the elements affects the band gap of our semiconductor. So if we have a higher band gap this means that we have a less electrical conductivity. And if we have a lower band gap that we have a more electrical conductivity. So the relation between our change in electro negativity or difference in electro negativity and the band gap is so let's go ahead and write this relation out is that if our band gap increases than our difference in electro negativity will also increase. So these electro negativity values is something that I have. But you can also look at your textbook or it can be even provided to you by your professor. So for indian which is I. N. Has a value of 1.7 Gallium. So G.A. a value of 1.6 and phosphorus OP has a value of 2.1. Let's go ahead and calculate these differences in electron negativity. So the delta E. N. Between our indium and phosphorus is 2.1 minus 1.7. And that difference is 0.4. Now the delta E. N. Between the gallium and our phosphorus that's 2.1 minus R 1.6 which gives us a difference of 0.5. So the change or the difference between electro negativity is between indian and phosphorus is smaller. So this means that there's a higher proportion of indium in the semiconductor, which leads to a lower band gap. So let's go ahead and arrange these semiconductors from the least proportional of indian to the highest. So the first one then Would be IN 0.25 and gallium 0.75. Then Indium 0. With Gallium 0.5 and phosphorus and last C Indium 0. gallium gallium 0.30 and phosphorus. So this right here is going to be my final answer for this problem. Thank you all so much for watching.