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Due to the presence everywhere of the cosmic background
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- 1 (a) Show that the entropy per photon in blackbody radiation is independent of the temperature, and in d spatial dimensions is given by En-d-1 s = (d + 1) n=1 E n-d n=1 (b) Show that the answer would have been d + 1 if the photons obeyed Boltzman statistics.arrow_forwardThe photons that make up the cosmic microwave background were emitted about 380,000 years after the Big Bang. Today, 13.8billion years after the Big Bang, the wavelengths of these photons have been stretched by a factor of about 1100 since they were emitted because lengths in the expanding universe have increased by that same factor of about 1100. Consider a cubical region of empty space in today’s universe 1.00 m on a side, with a volume of 1.00 m3. What was the length s0 of each side and the volume V0 of this same cubical region 380,000 years after the Big Bang? s0 = ? m V0 = ? m^3 Today the average density of ordinary matter in the universe is about 2.4×10−27 kg/m3. What was the average density ?(rho)0 of ordinary matter at the time that the photons in the cosmic microwave background radiation were emitted? (rho)0 = ? kg/m^3arrow_forwardThe photons that make up the cosmic microwave background were emitted about 380,000 years after the Big Bang. Today, 13.8 billion years after the Big Bang, the wavelengths of these photons have been stretched by a factor of about 1100 since they were emitted because lengths in the expanding universe have increased by that same factor of about 1100. Consider a cubical region of empty space in today's universe 1.00 m on a side, with a volume of 1.00 m³. What was the length so of each side and the volume V of this same cubical region 380,000 years after the Big Bang? So = Vo = Enter numeric value Today the average density of ordinary matter in the universe is about 2.4 × 10-27 kg/m³. What was the average density po of ordinary matter at the time that the photons in the cosmic microwave background radiation were emitted? Po = m m³ kg/m³arrow_forward
- Suppose your professor discovers dark matter, and it turns out to be a new type of subatomic particle which can share a state with one other particle – but no more – of the same type. In other words, the number of these (indistinguishable) particles in any given state can be 0, 1, or 2. (a) Derive the distribution function (analogous to the Bose-Einstein or Fermi-Dirac distribution) for the average occupancy of a state for particles of this type. It should be a function of e, µ, and kT. (b) What is the value of this distribution function at e = u? What value does the function approach for very large e? For very small e, under the assumption that u >> kT? (c) Suppose the "state" of the particle is defined only by its energy (do not consider spin, charge, etc.) Furthermore, suppose these particles are found in a system where the quantized energy levels are evenly spaced1 eV apart, and the lowest level has energy 0. What is the lowest possible energy of the system, if there are 10 of…arrow_forwardIn discussing molecular rotation, the quantum num- ber J is used rather than 7. Using the Boltzmann distribution, calculate nj/no for ¹H³5Cl for J = 0, 5, 10, and 20 at T = 1025 K. Does ny/no go through a maximum as J increases? If so, what can you say about the value of J corre- sponding to the maximum?arrow_forwardHow to solve this questionarrow_forward
- You are on an interstellar mission from the E arth to the 8.7 light-years distant star Sirius. Your spaceship can travel with 70% the spee d of light and has a cylindrical shape with a diameter of 6 m at the front surface and a le ngth of 25 m. You have to cross the interstel lar medium with an approximated density of 1 hydrogen atom/m3. (a) Calculate the time it takes your spaceshi p to reach Sirius. (b) Determine the mass of interstellar gas th at collides with your spaceship during the mi ssion.arrow_forwardUse Boltzmann distribution to solve this problem.A system consists of 3, 000 particles that can only occupy two energy levels: a nondegen-erate ground state of 0.052 eV and a threefold degenerate excited state at 0.156 eV. IfT = 900 K,(a) find the number of particles at each energy level.(b) what is the total energy of the system?arrow_forwardThe degeneracy pressure of the electrons can stabilize the collapse of a star due to gravity by equating the gravitational inward pressure with the outward electron gas degeneracy pressure. These cold stars called white dwarfs have small radii compared to their original size and this radius decreases as the original mass of the star increases. As the mass of a star increases, the electron energy increases to a point in which their energy has to be treated relativistically. (a) Evaluate the degeneracy pressure for ultra-relativistic electrons (problem 2, above). (b) As the pressure increases, the reaction e − + p → n + ν takes place. The neutrinos (ν) escape as matter is transparent to them, electrons and protons convert to neutrons until we are left with a neutron star. Using your expression for the degeneracy pressure obtained in (a) above, equate the gravitational pressureto the neutron degeneracy pressure (replace the electron’s mass in your degeneracy pressure expression from (a)…arrow_forward
- The Boltzmann constant has a value of kg = 0.69503476 cm-1/K. At what temperature would kBT be equivalent to the first energy level spacing, E₁-Eo ha for a harmonic oscillator of frequency w = 500 cm-1? This is a crude indication of the temperature needed to appreciably populate the n = 1 vibrational level. =arrow_forwardNeed help using Boltzmann's constant. If i'm trying to find average thermal velocity (Vrms) of a water droplet (spherical) and i'm given radius = 10^-6 m at 293.15 K, and the density of water is 0.997 g/mL (997 kg/m^3), how would i go about finding the Vrms? The droplet can also be treated as an ideal gas particle.arrow_forwardCould someone explain to me in detail why bringing a crystal substance to absolute zero isn't possible? I know it's not because of quantum mechanics and uncertainty like some people say, because particals at their lowest zero-point will have a temperature of exactly 0 K, even though they're still experiencing motion. From what I've gathered, the energy or time required to pull it off is infinite, but I can't find any equations or clear explanations as to why or how that is. And I also don't know if there's any other reasons beyond that. If you could give me a thourough a breakdown for how absolute zero is impossible as you possibly could, I'd greatly appreciate it. Take as much extra time as you need. As long as it's detailed and correct I'm happy. Though ideally I would before it come in before the end of the day.arrow_forward
- Modern PhysicsPhysicsISBN:9781111794378Author:Raymond A. Serway, Clement J. Moses, Curt A. MoyerPublisher:Cengage Learning