Chapter 9: Q16CQ (page 403)
Suppose that in Figure 9.27, the level labelled , rather than the one labelled, were metastable. Might the material still function as a laser? Explain.
Short Answer
Yes the material still function as laser .
Chapter 9: Q16CQ (page 403)
Suppose that in Figure 9.27, the level labelled , rather than the one labelled, were metastable. Might the material still function as a laser? Explain.
Yes the material still function as laser .
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The electrons’ contribution to the heat capacity of a metal is small and goes to as . We might try to calculate it via the total internal energy, localid="1660131882505" , but it is one of those integrals impossible to do in closed form, and localid="1660131274621" is the culprit. Still, we can explain why the heat capacity should go to zero and obtain a rough value.
(a) Starting withexpressed as in equation (34), show that the slope atis.
(b) Based on part (a), the accompanying figure is a good approximation to when is small. In a normal gas, such as air, whenis raised a little, all molecules, on average, gain a little energy, proportional to . Thus, the internal energy increases linearly with , and the heat capacity, , is roughly constant. Argue on the basis of the figure that in this fermion gas, as the temperature increases from to a small value , while some particles gain energy of roughly , not all do, and the number doing so is also roughly proportional to localid="1660131824460" . What effect does this have on the heat capacity?
(c)Viewing the total energy increase as simply = (number of particles whose energy increases) (energy change per particle) and assuming the density of states is simply a constant over the entire range of particle energies, show that the heat capacity under these lowest-temperature conditions should be proportional to . (Trying to be more precise is not really worthwhile, for the proportionality constant is subject to several corrections from effects we ignore).
By considering its constituents, determine the dimensions (e.g. length, distance over lime. etc.) of the denominator in equation. Why is the result sensible?
When a star has nearly bumped up its intimal fuel, it may become a white dwarf. It is crushed under its own enormous gravitational forces to the point at which the exclusion principle for the electrons becomes a factor. A smaller size would decrease the gravitational potential energy, but assuming the electrons to be packed into the lowest energy states consistent with the exclusion principle, "squeezing" the potential well necessarily increases the energies of all the electrons (by shortening their wavelengths). If gravitation and the electron exclusion principle are the only factors, there is minimum total energy and corresponding equilibrium radius.
(a) Treat the electrons in a white dwarf as a quantum gas. The minimum energy allowed by the exclusion principle (see Exercise 67) is
Note that as the volume Vis decreased, the energy does increase. For a neutral star. the number of electrons, N, equals the number of protons. If protons account for half of the white dwarf's mass M (neutrons accounting for the other half). Show that the minimum electron energy may be written
Where, R is the star's radius?
(b) The gravitational potential energy of a sphere of mass Mand radius Ris given by
Taking both factors into account, show that the minimum total energy occurs when
(c) Evaluate this radius for a star whose mass is equal to that of our Sun 2x1030kg.
(d) White dwarfs are comparable to the size of Earth. Does the value in part (c) agree?
In a large system of distinguishable harmonic oscillator how high does the temperature have to be for the probable number of particles occupying the ground state to be less than 1 ?
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