An electron confinedtoa cubic 3D infinite well 1 nu on aside.

  1. What are thethree lowest differentenergies?
  2. To how many different states do these three energies correspond?

Short Answer

Expert verified

1. The three lowest different energies are as given below.

E1,1,1=1.13cVE1,1,2=2.26eVE1,2,2=3.38eV

2. Among the first three energies computed in part (a), there are one ground state, three degenerate first excited states, and three degenerate second exciting levels for a total of seven states.

Step by step solution

01

A concept of energy:

The three quantum numbers describe the states of the cubic square well are (nx,ny,nz)and the energy for a given set of quantum numbers.

Then write the relation as below.

Enx,ny,nz=(nx2+ny2+nz2)π2h22mL2

02

(a) Determine the three lowest different energies:

For the ground state nxny,nz=1,1,1which has the lowest energy.

The corresponding energy of the ground state is,

E1,1,1=12+12+12π2h22mL2

Substitute 1.054×10-34J.s for h , 9.1094×10-31kg for m, and 1nm for L in the above equation.

E1,1,1=12+12+12π2h22mL2=3(3.14159)21.054×1034J.s229.1094×1031kg(1nm)2=3(3.14159)21.054×1034J.s229.1094×1031kg(1nm)109mnm2=36.0247×1020J

Thus, the lowest energy is,

E1,1,1=36.0247×10-20J=1.81×10-20J=1.81×10-20J1eV1.60218×1010J=1.13cV

There is only one such state.

03

Determine the next lowest energies:

The lowest energy correspond to any of the three choice of quantum numbers,

nxny,nz=2,1,1nxny,nz=1,2,1nxny,nz=1,1,2

All of which have the same energy as,

Enx,ny,nz=(nx2+ny2+nz2)π2h22mL2

Substitute 1.054×10-34J.s for h, 9.1094×10-31kg for m, and 1 nm for L in the above equation.

E1,1,2=12+12+22π2h22mL2=6(3.14159)21.054×1034J.s229.1094×1031kg(1nm)2=6(3.14159)21.054×1034J.s229.1094×1031kg(1nm)109mnm2=66.0247×1020JE1,1,2=3.62×1019JJ1.6×1010eV=2.26eV

04

Determine the next lowest energies:

The lowest energy corresponds to any of the three choice of quantum numbers,

nxny,nz=1,2,2nxny,nz=2,1,2nxny,nz=2,2,1

All of which have the same energy as,

Enx,ny,nz=(nx2+ny2+nz2)π2h22mL2Enx,ny,nz=(nx2+ny2+nz2)π2h22mL2

Substitute 1.054×10-34J.s for h , 9.1094×10-31kg for m , and 1 nm for L in the above equation.

E1,2,2=12+22+22π2h22mL2=9(3.14159)21.054×1034J.s229.1094×1031kg(1nm)2=9(3.14159)21.054×1034J.s229.1094×1031kg(1nm)109mnm2=96.0247×1020JE1,2,2=5.42×1019JeV1.6×1019J=3.38eV

05

(b) Determine different states do these three energies correspond:

Among the first three energies computed in part (a), there are one ground state, three degenerate first excited states, and three degenerate second excited levels for a total of seven states.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

We have noted that for a given energy, as lincreases, the motion is more like a circle at a constant radius, with the rotational energy increasing as the radial energy correspondingly decreases. But is the radial kinetic energy 0 for the largest lvalues? Calculate the ratio of expectation values, radial energy to rotational energy, for the(n,l,mt)=(2.1,+1)state. Use the operators

KErad=-h22m1r2r(rr)KErad=h2l(l+1)2mr2

Which we deduce from equation (7-30).

Question: Section 7.5 argues that knowing all three components of would violate the uncertainty principle. Knowing its magnitude and one component does not. What about knowing its magnitude and two components? Would be left any freedom at all and if so, do you think it would be enough to satisfy the uncertainly principle?

For an electron in the(n,l,ml)=(2,0,0) state in a hydrogen atom, (a) write the solution of the time-independent Schrodinger equation,

(b) verify explicitly that it is a solution with the expected angular momentum and energy.

Consider a cubic 3D infinite well.

(a) How many different wave functions have the same energy as the one for which (nx,ny,nz)=(5,1,1)?

(b) Into how many different energy levels would this level split if the length of one side were increased by 5% ?

(c) Make a scale diagram, similar to Figure 3, illustrating the energy splitting of the previously degenerate wave functions.

(d) Is there any degeneracy left? If so, how might it be “destroyed”?

In Appendix G. the operator for the square of the angular momentum is shown to be

L^2=-h2[cscθθsinθθ+csc2θ2ϕ2]

Use this to rewrite equation (7-19) asL^2Φ=-Ch2Φ

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free