A photon of green light has a wavelength of 520 nm. Find the phot...

Question

A photon of green light has a wavelength of 520 nm. Find the photon's frequency, magnitude of momentum, and energy. Express the energy in both joules and electron volts.

Accepted Answer

Hi everyone. In this practice problem, we're being asked to calculate the frequency momentum and energy of a fluorescent lamp emitting red light of wave flang 640 nanometer. So for a single photon in the light, we're being asked to determine first the photons momentum, its energy in Jos as well as in EV and also its frequency. So the options given are listing the different frequencies, momentum and energies both in Jos and EV. Awesome. So we have the formula of C which is the speed of light to be equals to f multiplied by LAMBDA where F is the frequency of the photon lambda is the wavelength and C is the speed of light or velocity of light where C is just a constant of 3.00 times 10 to the power of 8 m per second. So in this case, in order for us to get our frequency, which is the first part of this question, we can rearrange our equation so that we get frequency to then be equals to C divided by lambda. We were given the lambda to a wavelength value in the problem statement. So we can actually substitute our values to get our frequency directly. So frequency will then equals to C which is 3.00 times 10 to the power of 8 m per second divided by lambda. And the lambda is given in nanometer. So we have to multiply our 640 with 10 to the power of negative nine to convert it from nanometer to meter. Calculating this, we can then get our frequency value to then be equal to 4.69 times 10 to the power of 14, 1 over seconds or one divided by seconds or Hertz. So that's the first part of this problem statement which is the frequency. So next, what we want to calculate is the momentum. In order to calculate the momentum, we want to recall the equation for momentum or B which will be equals to H divided by a lambda. H is just going to be the planks constant which is a constant of 6.63 times 10 to the power of negative 34 seconds. Lambda is still our lambda that is given in the problem statement. So we can actually calculate our momentum right away. So momentum will then be equals to 6.63 times 10 to the power of negative 34 2 seconds. Divide that by a lambda which is going to then be equals to 640 times 10 to the power of negative 9 m because we want it in meter. And in calculating this, we will get our um momentum to then be equals to 1.04 times 10 to the power of negative 27 kg meter per seconds. So now that we have our frequency and our momentum, the only thing that's left is just the energy of the photon. So the energy of the photon can be calculated by multiplying the momentum with its or the with with the speed of light or sea. So in this case, we can directly calculate this. So the momentum we found that previously to be 1.04 times 10 to the power of negative 27 kg meter per seconds multiplied that by the speed of light which is 3.00 times 10 to the power of eight meters per second. And then calculating that we will get the energy in Jules which is going to be 3.12 times 10 to the power of negative 19 Jules. We are also asked the energy in ev so we wanna convert the energy to ev so we have 3.12 times 10 to the power of negative 19 Jules. And in order for us to convert this into ev, we wanna multiply this with one ev for every or one ev and divide it by 1.602 times 10 to the power of negative 19 jos. So one ev will equals to 1.602 times 10 to the power of negative 19 Jules, which is why we are dividing it by that calculating this, we will then get our energy to then be equals to 1.95 ev. So there we go. There we have all of the answer to this problem statement. First, we have the frequency which is going to equal to four or 0.69 times 10 to the power of 14 Hertz. And then we have the momentum which is one point over times 10 to the power of negative 27 kg meters divided by seconds. And then the energy in Jules which is 3.12 times 10 to the power of negative 19 Joles and the energy in EV which is 1.95 EV. And all of that will then correspond to option see in our answer choices. So option C will be the answer to this particular practice problem and that will be it for this video. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on our website and that will be it for this video. Thank you.

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A photon of green light has a wavelength of 520 nm. Find the photon's frequency, magnitude of momentum, and energy. Express the energy in both joules and electron volts.

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