A photovoltaic cell contains a p–n junction that that converts
s...

Question

A photovoltaic cell contains a p–n junction that that converts 
solar light to electricity. An optimum semiconductor 
would have its band-gap energy matched to the wavelength 
of maximum solar intensity at the Earth’s surface. 
(a) What is the color and approximate wavelength of maximum 
solar intensity at the Earth’s surface? Refer to the 
figure for Problem 12.102.

Accepted Answer

welcome back everyone In this example, we have radiation with a wavelength of 8.5 nanometers emitted by a celestial body, in which region of the electromagnetic spectrum is the radiation located. And if an observatory detects 5.2 times 10 to the eighth power photons per second, calculate the total energy that is detected in one hour. So our first step is to recognize that because the prompt mentions wavelength recalled, that wavelength is represented by the symbol lambda, which is typically expressed in units of meters. However, the prompt gives us our wavelength in nanometers. So we're going to begin by converting 8.5 nanometers to the proper unit meters. By recalling that our prefix nano tells us that we have 10 to the negative ninth power of our base unit meter equivalent to one nanometer. This allows us to cancel our units of nanometers, leaving us with meters as our final unit for wavelength. And this is going to give us 8.5 times 10 to the negative ninth power meters as our wavelength. Now we want to go ahead and see where on the electromagnetic spectrum this corresponds. So I'm going to paste the electromagnetic spectrum down below. So based on our calculation, we see that we have a power of 10 to the negative ninth power. And looking at our electromagnetic spectrum, we see our wavelength is represented by lambda here on the bottom and we can find a power of 10 to the eighth power here where we have the beginning of our UV light and the ending of our X rays. And so we have 10 to the eighth power here at this point where we have 10 to the negative 10th power here at this point. And so we can assume that 10 to the negative ninth power will lie between that corresponding to x rays as the position of our wavelength calculation on the spectrum. And so we can say that this corresponds to x rays. So this would be our first answer. And next we want to go ahead and determine the total energy detected In one hour. So we want to recall our formula for the energy of a photon and we can recall that energy is equal to Planck's constant represented by H. So we'll use a different color here. We'll use the color purple. This is then multiplied by our speed of light which will represent in blue. And then this is divided by our wavelength which will represent by lambda and red. So we're gonna plug in what we know into this formula to solve for energy and this is specifically energy of one photon. So this is going to be equal to Planck's constant. Which we should recall is 6.626 times 10 to the negative 34th power units of joules times seconds, which is then multiplied by our speed of light, which we should recall is 3.0 times 10 to the eighth power units of meters per second. And then we're dividing by our wavelength which above we calculated to be 8.5 times 10 to the negative ninth power meters. So canceling out our units, we can get rid of meters with meters in the numerator. We can also get rid of seconds with seconds in the denominator here, leaving us with jewels as our final unit for energy. And so in our calculators, we're going to plug in our quotient and this is going to yield a result of 2.338, 6 times 10 to the negative 17th power. And we have units of jewels and this is per photon. So because we need to determine the total energy emitted in one hour, we're going to go ahead and use the information from the prompt in a dimensional analysis step. So we can say that our total energy Is equal to from the prompt, we're told that the observatory detects 5.2 times 10 to the 8th power photons per second. So we'll have per second in the denominator and we're going to multiply this by our calculation for the energy of each photon. And so what we want to do is cancel out our units of photon. So we're going to have to 0.3386 times 10. We're sorry we're going to have one in our denominator. So we would have for one photon. We understand that that is equivalent to our energy of 2.3386 times 10 to the negative 17th power jewels. So we can so far cancel our units of photons now we have jewels per second but we just want jewels per hour as our final unit. So we're going to multiply by our next conversion factor where we're going to recall that we have In 60 seconds an equivalent of one minute. Now we can cancel out seconds and focus on going from minutes to hours by recalling our last conversion factor where we have for minutes an equivalent of one hour. So this allows us to cancel out minutes leaving us with our final units of joules per hour which is what the prompt states to calculate. And so in our calculators, when we plug in our quotients we're going to yield a result from these products of 4.4 times 10 to the negative fifth power with units of joules per hour. And so this is the total energy that is detected from the observatory per hour of all of the photons emitted from our celestial body. And this is going to be our second and final answer to complete this example. 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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