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13) Jo The left-hand side is with r2 = x\ + . . 14) where eq. 5) has been used. The right-hand side of eq. 13) is with y = r2, dr = dy/2& r°° Cdd\ 0 D Equating eqs. 1. ) Ge (1) Main energy gap at 300 K (eV) (2) Effective densities of states (cm~3) at 300 K (see eqs. 9)) (3) Effective electron masses in units of the electron rest mass (see eq. 9)) at4K Parabolic mass near band bottom u (4) Effective hole masses in units of the electron rest mass at 4 K heavy-hole band (mh) light-hole band (mj split-off band (ms) (5) Density-of-states effective masses at 4 K in units of the electron rest mass (see eqs.
One can simply sum over all permitted integers N19 N2... etc. instead of summing over all N and all i, since this procedure covers all possible states of the system if the particles involved are indistinguishable. For simplicity we shall consider a system of two single-particle states. Thus eq. 2) becomes, using only simple algebra, The upper limits in the summations are either unity, so that a quantum state is either empty or full, or infinity, so that a quantum state can accommodate any number of particles.
One need not appeal to eq. 13) if one observes that in thermal equilibrium V|a = 0 from thermodynamics, that j = 0 by hypothesis, so that the coefficient of Vv in eq. 12) must vanish. The relation obtained may then be expected to be valid also away from, but in the neighborhood of, equilibrium. ' = Ahexp(-\i/kT) \q\D so that one has —,— = 1 vkT The Einstein ratio \q\ D/v is therefore just the typical thermal energy kT of a carrier in this simple case. 11). 4]. 6] for historical reviews). For large departures from equilibrium, however, eqs.