Problem

(a) What Hubble constant corresponds to an approx…

Problem 14 Hard Difficulty

The peak intensity of the CMBR occurs at a wavelength of 1.1 $\mathrm{mm}$ (a) What is the energy in eV of a 1.1 -mm photon? (b) There are approximately $10^{9}$ photons for each massive particle in deep space. Calculate the energy of $10^{9}$ such photons. (c) If the average massive particle in space has a
mass half that of a proton, what energy would be created by converting its mass to energy? (d) Does this imply that space is "matter dominated"? Explain briefly.

Answer

a. $1.1 \times 10^{-3} \mathrm{eV}$
b. 1.1 MeV
c. $4.69 \times 10^{2} \mathrm{MeV}$
d. Yes, the space is matter dominated as huge amount of energy is associated with the mass of massive particles and there subsequent particles which is obvious from the above results. The energy by converting mass to energy is almost 500 times the energy from the photon of 1.1 $\mathrm{mm}$ .

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Chapter 34

Frontiers of Physics

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Video Transcript

in the first part of the problem, we need to calculate the energy associate ID to a foot on who's Vaillant is 1.1 millimeter. So that's still be It went now. Energy is given by H Timesnew. Where new is the frequency and h is the planks constant? One can convert the frequency in terms off speed of light and lambda and you get HC over. Lander, this is the formula that we're gonna use. So we know that energy is H c o ver lander. I mean slug in all the values. So the text book tells us the value of age or the planks. Constant, an MTV is this speed of light is three times 10 to the eight meters per second. And Linda is 1.1 millimeter and one million. We test into the minus three meter. Okay, remember to do all the calculations and s I units. And once you sold it, you get the energy to be 1.1 times 10 to the minus three electron volts. So the energy off the photon off we've let 1.1 millimeter. Is this good? Now for the second part of the problem, is how much energy is contained in 10 to the nine such four tones. So okay, since one Ford on contains this much amount of energy, tend to the 94 dawns would contain tender the nine times this. Right. So all I do is multi play. Tend to the nine to the energy off, one for dawn, which is roughly 1.1 times 10 to the minus three electron volts. Yeah. So this gives 1.1 times 10 to the six electrons rules. So this I can write us one mega elect 1.1 mega electron told Megan Means tend to the six. Good. All right, so we've done the 1st 2 but for the third box, you're given that the average mass off a massive particle is half the mass of a proton so mass off a particle is half the mass off proton find its energy. So and their G we know from Einstein's relation is M C Square and M is given to be half the mass of Proton C is the speed of light. So again we will plug in the massive proton first, which is in electron volts, because all of her calculation has used the units off electron roll, so we will keep that uniform. So it's this in electron bulls time C Square now see is the same that we used before, which is three times 10 to the eight meters per second. On solving this, you get five times 10 to 2 mega electron novels. Now mega means tend to the six. So I'll explicit. Read. Write this in terms of electron rules. So either I can say I have 500 mega balls or I can say I have five times 10 to the eighth electron balls. It's the same thing now we need to see. We need to compare the two energies. So if you compare the two energies so the energy off photons is 1.1 of maybe and the energy of the massive particle is 500 Emery, So definitely them. The energy of the massive particles is more so. The energy is dominated by the massive particles on by how much? 500 times. And it's massive particle dominated because the energy of massive particle is more than the energy off the photons. All right, All the calculations have been done in the units off electron world

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