Suppose we have a system of identical particles moving in just one dimension and for which the energy quantization relationship isE=bn2/3, wherebis a constant andan integer quantum number. Discuss whether the density of states should be independent ofE, an increasing function ofE, or a decreasing function ofE.

Short Answer

Expert verified

Based on our above expression for D(E), the density of states is thus an increasing function ofE . More accurately, the density of states is directly proportional to E1/2.

Step by step solution

01

Step 1:

In this problem, we are given the relationship between the energyEand the quantum staten:

E=bn2/3,

Whereb is constant. Our task is to determine whether the density of states D(E)is increasing, decreasing, or independent of E.

02

express  n in terms of E

Given the quantized energy equation above, let us first express nin terms of E:

E=bn2/3n2/3=Ebn=(Eb)3/2

03

expression for the density of states D(E)

Now that we have nin terms ofE, let us take the derivative of both sides:

d(n)=d(Eb)3/2dn=32E1/2b3/2dE=32Eb3dE

Dividing both sides by dEwe obtain the expression for the density of states D(E)associated with the energy relation above:

dndE=32Eb3dEdE=32Eb3D(E)=32Eb3

04

density is a increasing function

Based on our above expression for D(E), the density of states is thus an increasing function of E. More accurately, the density of states is directly proportional to E1/2.

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Most popular questions from this chapter

Show that in the limithω0kBT. Equation (9.15) becomes (9.28).

The electrons’ contribution to the heat capacity of a metal is small and goes to 0as T0. We might try to calculate it via the total internal energy, localid="1660131882505" U=EN(E)D(E)dE, but it is one of those integrals impossible to do in closed form, and localid="1660131274621" N(E)FDis the culprit. Still, we can explain why the heat capacity should go to zero and obtain a rough value.

(a) Starting withN(E)FDexpressed as in equation (34), show that the slope N(E)FDdEatE=EFis-1(4kBT).

(b) Based on part (a), the accompanying figure is a good approximation to N(E)FDwhen Tis small. In a normal gas, such as air, whenTis raised a little, all molecules, on average, gain a little energy, proportional to kBT. Thus, the internal energy Uincreases linearly with T, and the heat capacity, UT, is roughly constant. Argue on the basis of the figure that in this fermion gas, as the temperature increases from 0to a small value T, while some particles gain energy of roughly kBT, not all do, and the number doing so is also roughly proportional to localid="1660131824460" T. What effect does this have on the heat capacity?

(c)Viewing the total energy increase as simply U= (number of particles whose energy increases) (energy change per particle) and assuming the density of states is simply a constant Dover the entire range of particle energies, show that the heat capacity under these lowest-temperature conditions should be proportional to kBREFT. (Trying to be more precise is not really worthwhile, for the proportionality constant is subject to several corrections from effects we ignore).

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