Question

Light consists of photons: noninteracting quantum oscillator particles with an energy E-hv- e, where is their frequency, A is their wavelength, and c is their speed (ie the speed of light) The momentum of a photon, p has a special definition, because photons have no mass. We instead use the De Broglie identity p-. This means that we can relate the "wavevector" from the solution to the Schrödinger equation to the momentum p - hk. a) [2 marks] Starting from the general expression for the density of states for a quantum gas in a volume V: dk 2 Obtain an expression for the density of states for photons in terms of energy, g (s) de b) [3 marks] Use the expression for the density of states you found above, derive an expression for the number density of photons per unit volume, p (E) de, by using an appropriate distribution function. c) [5 marks] In general, the number of photons is never conserved - they are created and destroyed with no energy penalty, and thus we may set the chemical potential μ-0. Using this fact, derive an expression for the total energy density of the photon gas, using the expression you obtained in b) You may find the following relation useful: 15

please do the full working everyone is just jumping to the answer im trying to understand the working please

Answer 1

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