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An electron (mass m) is contained in a cubical box of widths Lx=Ly=Lz. (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 ofh/8mL2 gives 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 7 different frequencies.

(b) The multiple of h8mL2is 1.

(c) The multiple ofh8mL2 is 2.

(d) The multiple ofh8mL2 is 3.

(e) The multiple ofh8mL2 is 9.

(f) The multiple ofh8mL2 is 8.

(f) The multiple ofh8mL2 is 6.

Step by step solution

01

The energy of three dimensional electron traps:

The energy equation for an electron trapped in a three-dimensional cell is given by,

Enx,ny,nz=h28m(nx2Lx2+ny2Ly2+nz2Lz2) ….. (1)

Here,nx is quantum number for which the electron’s matter wave fits in well width Lx,ny is quantum number for which the electron’s matter wave fits in well width Lx,nz is quantum number for which the electron’s matter wave fits in well width Lz, h is plank constant, and m is mass of the electron.

The frequency in terms of the energy is given by,

f=Eh ….. (2)

02

(a) Find the number of different frequencies of light that can electron emit or absorb:

The energies of lowest five levels are as follows.

Enx,ny,nz=h28m(nx2Lx2+ny2Ly2+nz2Lz2)

Substitute L forLx,LforLy,LforLz,1fornx,1forny,and 1 fornzin equation (1).

E1,1,1=h28m1L2+1L2+1L2=3h28mL2

And similarly,

E2,1,1=6h28mE2,2,1=9h28mL2E1,1,3=11h28mL2E2,2,2=12h28mL2

From these energies, it can be seen that the non-iterated differences in the energy occurs between:

  • The fourth excited state and the ground state
  • The third excited state and the ground state
  • The third excited state and the first excited state
  • The fourth excited state and the third excited state
  • The third excited state and the second excited state
  • The second excited state and the ground state
  • The second excited state and the first excited state

Therefore, there are 7 different frequencies.

03

(b) Find the multiple of h/8mL2 for the lowest frequency:

The transition between the energies above is simply the differences between the factors that are multiplied by h28mL2.

The lowest frequency occur at (E1,1,3E2,2,2).

Find the difference betweenE1,1,3 and E2,2,2.

E1,1,3-E2,2,2=12h28mL2-11h28mL2=h28mL2

Find the frequency by using equation (2).

f1,1,32,2,2,=h28mL2h=h28mL2

Therefore, the multiple ofh28mL2 is 1.

04

(c) Find the multiple of for the second lowest frequency:

The second lowest frequency occur at (E1,1,3E2,2,1).

Find the difference betweenE1,1,3 and E2,2,1.

E1,1,3-E2,2,1=11h28mL2-9h28mL2=2h28mL2

Find the frequency by using equation (2).

E1,1,32,2,1=2h28mL2h1=2h28mL2

Therefore, the multiple ofh28mL2 is 2.

05

(d) Find the multiple of h/8mL2 for the third lowest frequency:

The third lowest frequency occur at (E2,1,1E2,2,1,).

Find the difference between E2,1,1,and E2,2,1.

E2,2,1-E2,1,1=9h28mL2-6h28mL2=3h28mL2

Find the frequency by using equation (2).

f2,1,12,2,1=3h28mL2h1=3h28mL2

Therefore, the multiple ofh28mL2is 3.

06

(e) Find the multiple of h/8mL2 for the highest frequency:

The third lowest frequency occur at (E1,1,1E2,2,2).

Find the difference betweenE1,1,1and E2,,2,2.

E2,2,2-E1,1,1=12h28mL2-3h28mL2=9h28mL2

Find the frequency by using equation (2).

f1,1,12,2,2=9h28mL2h=9h28mL2

Therefore, the multiple ofh28mL2is 9.

07

(f) Find the multiple of h/8mL2 for the second highest frequency:

The third lowest frequency occur at (E1,1,1E1,1,3).

Find the difference betweenE1,1,1 and E1,1,3.

role="math" localid="1661940120369" E1,1,3-E1,1,1=11h28mL2-3h28mL2=8h28mL2

Find the frequency by using equation (2).

f1,1,11,1,3=8h28mL2h=8h28mL2

Therefore, the multiple ofh28mL2 is 8.

08

(g) Find the multiple of h/8mL2 for the third highest frequency:

The third lowest frequency occur at (E1,1,1E2,2,1).

Find the difference betweenE1,1,1 and E2,2,1.

E2,2,1-E1,1,1=9h28mL2-3h28mL2=6h28mL2

Find the frequency by using equation (2).

f1,1,12,2,1=6h28mL2h=6h28mL2

Therefore, the multiple ofh28mL2 is 6.

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