A diode laser emits at a wavelength of 987 nm. (a) In what
porti...

Question

A diode laser emits at a wavelength of 987 nm. (a) In what 
portion of the electromagnetic spectrum is this radiation 
found? (b) All of its output energy is absorbed in a detector 
that measures a total energy of 0.52 J over a period of 32 s. 
How many photons per second are being emitted by the 
laser?

Accepted Answer

Welcome back everyone in this example. We have an injection laser diode which radiates light with a wavelength of 325 nanometers. Where does this radiation fall on the electromagnetic spectrum and all the energy output of the injection laser diode was measured over a period of 92 seconds and the total energy recorded is 920.24 jules. We need to calculate the number of photons emitted per second. So our first part is to figure out where on the electromagnetic spectrum this wavelength of light occurs at. And we're told that we have the wavelength of 325 nanometers. But we want to recall that when it comes to wavelength, which we recall is represented by the symbol lambda. This is expressed in units of meters. And so we're going to take the given wavelength 325 nanometers. And we want to convert from nanometers in the denominator, two m in the numerator. And we should recall that our prefix nano tells us that we have 10 to the negative ninth power of our base unit meter for one nanometer. Now we're able to cancel out our units of nanometers, leaving us with meters as our final unit for wavelength. And we're going to get a value of 3.25 times 10 to the negative seventh power meters as our wavelength. Now I'm gonna paste in an image of our electromagnetic spectrum so we can see what sort of light this wavelength falls on. So here we have our electromagnetic spectrum and I'm just going to make this a bit larger and we can find our reading for wavelength again by our symbol lambda. So looking at our calculation, we have a power of 10 to the negative seventh meters. And we can see that on our electromagnetic spectrum. We have a power of 10 to the negative eighth meters in wavelength and 10 to the negative six power meters in wavelength here. And actually let's make this a bit bigger. So with this size it should be a bit more visible. Again, we see we have 10 to the eighth power meters of wavelength in this region where we have 10 to the six power meters of wavelength in this region. And so that would mean that we can assume that 10 to the negative seventh power meters of wavelength is just at the beginning where our visible light spectrum is starting and where the ultraviolet light spectrum is ending. And so we can say that and we'll just draw the symbol in the color pink here, that this is about 10 to the negative seven power meters of our wavelength. And so we can say that this is ultimately going to correspond to UV light due to the fact that again, as we stated, this is where UV light is ending in the visible spectrum is beginning right at 10 to the negative seven power nanometers. So now that we understand our first answer. So we'll highlight this because it's our first answer here. We need to determine the number of photons that is emitted based on our energy output of our laser. So we're gonna scroll down for more room and we want to recall our formula to solve for the energy of a photon, which we would recall is equal to Planck's constant, multiplied by our speed of light, which is then divided by our wavelength lambda. And once we find the energy of a photon, we can use that value to then determine how many photons are emitted by using the information on the energy and time that was measured from our injection laser diode. So we're gonna go ahead and plug in what we know for this formula. And we should recall that plank's constant. And we'll actually write this below here. So recall that plank's constant is equal to the value of 6.626 times 10 to the negative 34th power. And sorry, that's a three. So to the negative 34th power with units of jewels, times seconds. We then want to multiply by our speed of light which we should recall is equal to 3.0 times 10 to the eighth Power meters per second. And then in our denominator we're dividing by our wavelength which we calculated above for our injection laser as 3.25 times to the negative seventh power. And we converted two units of meters. So now we're able to cancel our units of meters in the numerator. With meters in the denominator here, we can also get rid of seconds in the numerator and seconds in the denominator and we're left with units of jewels as our final unit of energy. So this is going to yield us a result for the energy of our photon equal to 6.1163 times 10 to the negative 19th power. Sorry, that's a nine, so times 10 to the negative 19th power. And we have units of jewels and we're going to interpret this as jewels per photon since this is the energy of one photon. So now to get our final answer, we're going to undergo a dimensional analysis step where we're going to recall from the prompt that we measured 0.20 for jewels of energy that was emitted during 92 seconds of from that laser injection. And so we're going to use this as a conversion factor to multiply by what we just calculated where we want our Energy per photon and jewels in the denominator. So we'll plug in 6.113 or sorry 6.116, 3 Times 10 to the negative 19th power jewels. And then in our numerator we're going to have per photon. So we have one photon because that is the energy of one photon. And now this allows us to cancel out our units of jewels. Now we're left with photons per second. But we want to cancel out where actually we don't want to cancel this out because it would make sense to have photons per second to determine how many photons are emitted per second of the energy output of our injection laser. And so we can just plug everything into our calculators to get the value for The product of these two quotients equal to 4.3 times 10 to the 15th power. And we have units of photons per second. And so this is going to be our second and final answer as our photons per second that are emitted during the injection laser diode output. And so everything highlighted in yellow represents our two final answers. To complete this example. I hope that everything I reviewed was clear. If you have any questions, leave them down below and I'll see everyone in the next practice video.

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