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For a particle in a one-dimensional (ID) box, Enis proportional to a single quantum number n. Let us simplify things by ignoring the proportionality factor: En=n2 . For a 3D box, Enx,ny,nz=nx2+ny2+nz2, and the 2D box is fairly obvious.

(a) The table shows a start on accounting for allowed states. Complete the table, stopping after the 10th state (state, not energy) for all three cases.

(b) Find the number of states per energy difference for the first five states and the last five states for all three cases. For instance, for the first five in the ID case, it is 5 states per energy difference of 24, or5/24 .

(c) Overlooking the obviously crude aspects of this accounting, does the "density of states" seem to increase with energy, decrease with energy, or stay about the same?

Short Answer

Expert verified

(a). Complete table is given in the solution

(b). The number of states for first five and last five in1Dare524and564respectively, in 2D are58and57respectively and in 3D are56and52respectively.

c). The density of the states tends to decrease with the energy.

Step by step solution

01

Formula used 

The expression of number of states is given by,

N=nE2−E1

02

Complete the table 

a)

Given information is

The En is proportional to single quantum number n.

To complete the table up to 10th state, n should varies such that it gives the lowest amount of energy. The complete table is

role="math" localid="1660026759024" 1D
2D
3D

States (n)

Energy (n2)

States (nx,ny)

Energy (nx2+ny2)

States

(nx,ny,nz)

Energy

(nx2+ny2+nz2)

1
1
1,1
2
1,1,1
3

2

4

1,2

5

1,1,2

6

3

9

2,1

5

1,2,1

6

4

16

2,2

8

2,1,1

6

5

25

1,3

10

1,2,2

9

6

36

3,1

10

2,1,2

role="math" localid="1660028505713" 9

7

49

2,3

13

2,2,1

9

8

64

3,2

13

1,1,3

11

9

81

1,4

17

1,3,1

11

10

100

4,1

17

3,1,1

11

03

Calculate the number of states per energy difference.

(b)

The expression of number of states is given by,

N=nE2−E1

The number of states for first five in 1D is calculated as,

N=nE2−E1N(firstfive)=525−1=524

The number of states for last five in 1Dis calculated as,

N=nE2−E1N(lastfive)=5100−36=564

The number of states for first five in 2Dis calculated as,

N=nE2−E1

N(firstfive)=510−2=58

The number of states for last five in 2Dis calculated as,

N=nE2−E1N(lastfive)=517−10=57

The number of states for first five in 3Dis calculated as,

N=nE2−E1N(firstfive)=59−3=56

The number of states for last five in 3Dis calculated as,

N=nE2−E1N(lastfive)=511−9=52

Therefore, the number of states for first five and last five in 1D are 524 and 564 respectively, in 2D are 58 and 57 respectively and in 3D are 56 and 52 respectively.

04

The change in density of states with the energy

c)

From part (b), the density of the states tends to increase with the dimensionality but decrease with the energy.

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Most popular questions from this chapter

You have six shelves, one above the other and all above the floor, and six volumes of an encyclopedia, A, B, C, D, E and F.

(a) list all the ways you can arrange the volumes with five on the floor and one on the sixth/top shelf. One way might be(ABCDE_,_,_,_,_F).

(b) List all the ways you can arrange them with four on the floor and two on the third shelf.

(c) Show that there are many more ways, relative to pans (a) and (b), to arrange the six volumes with two on the floor and two each on the first and second shelves. (There are several ways to answer

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(I) Use these ideas to argue that the relative probabilities of occupying the lowest energy states should be higher for hosons than for classically distinguishable particles.

(g) Combine these ideas with a famous principle to argue that the relative probabilities of occupying the lowest states should he lower for fermions than for classically distinguishable particles.

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You will need the following integral:∫0∞(Be2±1)−1dz=±ln(1±1B).

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