Wide band-gap semiconductors have a band gap between
2 and 7 elec...

Question

Wide band-gap semiconductors have a band gap between
2 and 7 electron volts (eV), where 1 eV = 96.485 kJ>mol.
The wide band-gap semiconductor GaN, used to construct
the laser in Blu-ray DVD players, has a band gap of
3.44 eV. The material in the laser, GaxIn1-x N, has some
indium substituted for gallium.
(b) If the light from the device is blue, does partial substitution
of indium for gallium increase or decrease the band
gap of GaxIn1-x N compared to GaN?

Accepted Answer

Welcome back everyone. We're told that wide band gap semiconductors are materials which have a larger band gap than conventional semiconductors, usually in the range of 2 to 7 electra vaults where one electron volt is equal to 96.4 85 kg joules per mole. The wide band gap semiconductor, aluminum arsenic. Houston led has a band gap of 2.16 electro volts. It can be doped with gallium to form aluminum, aluminum gallium arsenic which has a band gap of 1.72 electro volts. We need to identify the types of electromagnetic radiation emitted by aluminum arsenic and aluminum gallium arsenic. So because we are dealing with defining the types of electromagnetic radiation, we're going to be recalling our formula which relates the energy of a photon to the speed of light times Plank's constant. So the speed of light would be c, Plank's constant would be h divided by our wavelength of our radiation. Where if we identify the wavelength, we will then know our radiation based on where it corresponds to on our electromagnetic spectrum. So our first step is to use the values that they give us for our band gap of our semiconductors. First, beginning with aluminum arsenic were given a band gap of 2.16 electro volts and we want to recognize that the energy of our photon is in units of jewels per photon. So we want to end up with those units. And we're going to begin by getting rid of the electrical electrical unit by using the info from the prompt where they tell us that we have for one electron volt an equivalent of 96.4 85 kg joules times inverse moles. And so using that as a conversion factor, you can see we can now cancel out electro volts and now we want to get rid of kilo joules. So we're going to multiply by our next conversion factor to go from kilo jewels into jewels. And we're going to recall that our prefix kilo tells us that we have 10 to the third power of our base unit jewels. Now canceling out killer jewels, we want to get rid of inverse moles. So we're going to multiply by our next conversion factor using avocados number to go from or where we can introduce photon. So we would have 6.22 times 10 to the 23rd power avocados number of photons, We're gonna use this number to relate this to one mole. Where we can now cancel out inverse moles which is technically in the denominator with moles over here. And this leaves us with jewels per photon as our final unit for our energy emitted by the photons of our first semiconductor aluminum arsenic. This is going to yield the results of 3.463 point 4608 times 10 to the negative 19 power. And as we stated, we have units of joules per photon. Now we want to find that wavelength now that we know the energy emitted by these photons from our first semiconductor. And we're going to rearrange our formula so that we have wavelength isolated and set equal to Planck's constant, multiplied by the speed of light but divided by the energy emitted by our photons. And this means that what we can do is set it up so that we recall. That plank's constant is 6.626 times 10 to the negative 34th power jules, time seconds. We're then going to multiply by the speed of light, which we should recall is 3.0 times 10 to the eighth power. So 3.8 times 10 to the eighth power meters per second. Then divided by the energy emitted by our photon. Which we just calculated as 3.46 oh eight times 10 to the negative 19th power jewels per photon. And recall that wavelength is recorded in units and we'll write this below here were called, that wavelength is recorded in units of nanometers. So canceling out our unit below here, we'll see that we can get rid of jewels with jewels in the denominator as well as seconds in the denominator. With seconds in the numerator here. And this leaves us with technically meters as our final unit, since photons were just was just are derived unit from using avocados number. So now we're just going to carefully plug this into our calculators. And this is going to give us a wave length equal to a value of 5.74 times 10 to the negative seventh power units of meters. And now we're just going to convert two nanometers. So we're going to place meters in the denominator and nanometers in the numerator. Recall that our prefix nano tells us that we have 10 to the negative ninth power meters equal to one nanometer. And this allows us to cancel out meters. We're left with nanometers as our final unit for our wavelength and this is going to yield a wavelength of 574 nanometers. And so referring to our electromagnetic spectrum, we would see that this wavelength corresponds to the visible light region, particularly where we would see the color green. And this would represent our first part of our answer as the type of electromagnetic radiation emitted by our aluminum arsenic semiconductor. Now we need to figure out the type of radiation emitted by our aluminum x gallium one minus X arsenic semiconductor. So following the same process as before, We're going to find the energy emitted from the photons of that radiation from the semiconductor. Beginning with the info from the prompt where the band gap for aluminum gallium arsenic was 1.72 electro volts. And again, we want to end up with jules per photon as our final unit as the unit for the energy of our photons. So multiplying to first get rid of the electro volt unit. We use the conversion factor given from the prompt where one electron volt is equivalent to 96.4 85 killed jules, times inverse moles. And let's make sure this is written clearly. So inverse moles. This is then multiplied by our next conversion factor. Once we cancel out electro volts to go from killer joules in the denominator to Jewels and our numerator, where we would recall that our prefix kilo tells us that we have 10 to the third power of our base unit, jewels, canceling out killer jules. We're now going to focus on getting rid of our inverse small unit by incorporating avocados number where we can introduce photons as a unit. So again, avocados number is six point oh 22 times 10 to the 23rd power photons equivalent to one mole. And now we can cancel out inverse moles with moles in the numerator, leaving us with jewels per photons as our final unit. And calculating this energy emitted by these photons, we're going to have an energy equal to 2.7558 times 10 to the negative 19th power jewels per photon. And so this would then lead us into calculating our wavelength where again, isolating for wavelength from our photon energy formula above, we would say that wavelength is equal to Planck's constant. Multiplied by the speed of light divided by our energy emitted by our photons. And this is now then plugged in as 6.626 times 10 to the negative 34th power jewels times seconds multiplied by our speed of light three point oh times 10 to the eighth power meters per second. Which is then divided by our energy emitted by our photons, which we just found to be 2.7558 times 10 to the negative 19th power jewels per photon. And yet again canceling out our units. We're going to get rid of jewels, We're going to get rid of seconds and we're going to cancel out photons since it was just are derived unit. Leaving us with meters as our final unit. And this is going to yield a wavelength in meters equal to 7.21 times 10 to the negative seven power meters, which again, we want to convert two nanometers for our proper unit of wavelength. So having nanometers in the numerator, we would recall that we have 10 to the negative ninth power of our base unit, meters equal to one nanometer and canceling out meters. We're left with nanometers. This is going to give us our wavelength for our second semiconductor aluminum gallium arsenic equal to 721 nanometers. And when we look to our electromagnetic spectrum in our textbooks, we see that this corresponds to the visible light region but will radiate the color red. And so this would be our second final answer to complete this example. So we can confirm that our aluminum gallium arsenic semiconductor emits visible light in the red region, and our aluminum arsenic semiconductor emits visible light in the green region on our electromagnetic spectrum. So what's highlighted in yellow are our two final answers. I hope everything I explained was clear. If you have any questions, please leave them down below and this will correspond to choice B as the correct answer to complete this example officially. I hope everything I went through is clear and I will see everyone in the next practice video.

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