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An electron (mass m) is contained in a rectangular corral of widths Lx=Land LZ=2L. (a) How many different frequencies of light could the electron emit or absorb if it makes a transition between a pair of the lowest five energy levels? What multiple of h/8mL2gives the (b) lowest, (c) second lowest, (d) third lowest, (e) highest, (f) second highest, and (g) third highest frequency?

Short Answer

Expert verified

(a) There are eightpossible frequencies

(b) Lowest possible frequency =0.75

(c) Second lowest frequency=1

(d) Third lowest frequency=1.25

(e) Highest frequency=3.75

(f) Second highest frequency=3

(g) Third highest frequency=2.25

Step by step solution

01

Introduction

The electron is a subatomic particle whose electric charge is negative one elementary charge. Electrons belong to the first generation of the lepton particle family and are generally thought to be elementary particles because they have no known components or substructure.

02

Concept

The quantized energies for an electron trapped in a two dimensional infinite potential well that forms a rectangular corral are

Enx,ny=h28mnx2Lx2+ny2Ly2

Here,

nx=Quantum number for which the electron’s matter wave fits in well width Lx.

ny=Quantum number for which the electron’s matter wave fits in well width Ly.

03

Calculation

The energy level for an electron in the two-dimensional box,

Enx,ny=h28mnx2Lx2+ny2Ly2

Here,Ly=Land Lz=2L

Enx,ny=h28mnx2Lx2+ny2Ly2

So,

=h28mnx2Lx2+ny22L2=h28mnx2L2+ny24L2=h28mnx2+ny24

First five lowest possible energies:

Enx,ny==h28mL2nx2+ny24

CaseI: (1,1)

E1,1==h28mL212+124=1.25h28mL2

Case IV: (2,1)

E1,1==h28mL212+124=1.25h28mL222+124=4.25h28mL2

Case V: (2,1)

E2,1==h28mL222+124=4.25h28mL2

Case VI: (2,2)

E2,2==h28mL222+224=5h28mL2

When energy difference is more, then the frequency is high, where as when energy difference is less, then the frequency is less

04

(a) Determine the different frequencies of light could the electron emit

Different frequencies of light formed:

(E1,2E1,1):0.75(E1,3E1,1):2(E2,1E1,1):3(E2,2E1,1):3.75(E1,3E1,2):1.25(E2,1E1,2):2.25(E2,2E1,2):3(E2,1E1,2):1(E2,2E1,3):1.75(E2,2E2,1):0.75

There are eightpossible frequencies.

05

(b) Determine, the lowest possible frequency

Lowest possible frequency =0.75

06

(c) Determine, the second lowest frequency

Second lowest frequency=1

07

(d) Determine the third lowest frequency

Third lowest frequency=1.25

08

(e) Determine the highest frequency

Highest frequency=3.75

09

(f) Determine the Second highest frequency

Second highest frequency=3

10

(g) Determine the Third highest frequency

Third highest frequency= 2.25.

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Most popular questions from this chapter

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