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Integrated ConceptsA 1.00-fm photon has a wavelength short enough to detect some information about nuclei. (a) What is the photon momentum? (b) What is its energy in joules and MeV? (c) What is the (relativistic) velocity of an electron with the same momentum? (d) Calculate the electron’s kinetic energy.

a) $6.63 \times 10^{-19} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$b) $1.24 \times 10^{3} \mathrm{MeV}$c) 0.999$c$d) $2.55 \times 10^{5} \mathrm{eV}$

Physics 103

Chapter 29

Introduction to Quantum Physics

Quantum Physics

Rutgers, The State University of New Jersey

University of Washington

Hope College

McMaster University

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for this particular question, we are being asked to find the wavelength of a photon relating to the momentum of a proton moving at a portion of speed of light. Then we also are asked to find the energy of the photons in mega electron volts and the kinetic energy of the photo proton in mega electron volts. The first part part a calculate the wavelength of a photon that has the same momentum is a proton, maybe in 10% of the speed of like now, in order to find the wavelength, we need to know what the momentum of that proton would be. So the first thing we're gonna do is find the momentum of a proton moving at 10% of speed of light. Momentum is mass times velocity. So the momentum of proton is the mass, which is 1.67 times 10 to the negative 27 kilograms. And in this case, they want to know when it's moving at 1% point. A one is the decimal in the speed of light is three times 10 to the eighth meters per second. So this proton is going to have a momentum of 5.1 times 10 to the negative. 21st kilogram meters per second. Now, this isn't actually an answer. This is just something we need in order to solve again. The question states calculate the wavelength of a photon that has a particular momentum. Now that we're trying to find the wavelength, we know wavelength is plank's constant. Divided by momentum, we have both of these values now Plank's constant 6.626 times 10 to the negative 34th tool seconds divided by the moment when we found above and we get a wavelength of 1.323 times 10 to the negative 13th meters. Now, this problem didn't ask for it in centimeters or Nana meters. I'm just gonna leave it right as ISS The second part, what is the energy of the photons in mega electron volts on the scroll down Real quick, part B, The energy of the photons we know to find the energy we take the momentum and multiplied by the speed of like good news is we have both of these values because we found the momentum about 5.1 times 10 to the negative 21st kilograms meters per second and the speed of light is three times 10 to the eighth meters per second. Now, if I just saw that most by those two numbers together, I would get an answer in jewels. But I'm gonna continue below here into a little bit of conversions because we want an answering electron volts. So in my mind, we don't even need to find the intermediate answer. Quick conversion. We know that there is one electron able for every 1.6 times 10 to the negative 19th jewels. And there are there is one mega electron bold for every one times 10 to the sixth electron volts. What this does is Jules, was the answer. We had switched colors. When we must buy that the jewels cancel out. You look here, the electron volts cancel out and the unit that were left with will be mega electron bowls. And the answer to this part of the problem is nine 0.39 mega TV's. For the final part of the question. Find the kinetic energy kinetic energy of the proton, also in mega electron volts. 1/2 and he squared the mass of the proton is 1.67 times 10 to the negative. 27th The velocity. If we remember, it was 1% of the speed of light. So point a one three times 10 to the eighth gonna square that value that will give us and value in jewels. And if I used the same conversions above, I'm not gonna write those out at this time. But if I use that same conversion to get to Mega Electron volts, we will get an answer of 4.70 times 10 to the negative two mega electron volts. You don't want to write that with scientific notation. 0.0 for 70 mega electron volts would be would suffice. But you can very easily see that the photon has a lot more energy than the proton does.

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