How Does Quantum Tunneling Affect Electron Confinement in Potential Wells?

In summary, we discussed an electron with a total energy of Eo = 4.4 eV trapped in a potential well. We found the ratio of the wavelength in Region III to the wavelength in Region I to be λ III / λI = 1.77. We also determined that the maximum width of Region I, where there are no other nodes, is .29 nm. Finally, we calculated the minimum energy of the electron such that it can never escape the well in Region I to be 3 eV. The concept of the particle being trapped in Region I was also discussed.
  • #1
gv3
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Homework Statement


odd-well.jpg


An electron with a total energy of Eo = 4.4 eV is in the potential well shown above.

1) Find the ratio of the wavelength in Region III to the wavelength in Region I.

λ III / λI = 1.772) Given that the wave function of the electron vanishes at the left boundary of Region I, what is the maximum width dof Region I such that there are no other nodes in Region I?

.29 nm

3)What is the minimum energy of an electron such that it can never escape the well in Region I?

? eV

Homework Equations


E=hf
T= e-2bL
b=√(2m(U-E)/ħ2)
R= (K1-K2)2/(K1+K2)2

The Attempt at a Solution


Originally i thought that i need to set the probability of the particle transmitting equal to zero. But e can never be zero. So then i tried setting the probability of the particle reflecting equal to 1 but that still didn't lead to anything. Am i missing an equation?
 
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  • #2
Think about what it means for the particle to escape region I. Where would the particle be found then? What would be the energy of the particle after having escaped?
 
  • #3
DrClaude said:
Think about what it means for the particle to escape region I. Where would the particle be found then? What would be the energy of the particle after having escaped?
It would escape if region 2. Its energy would be -.6 correct?
 
  • #4
gv3 said:
It would escape if region 2. Its energy would be -.6 correct?
The lowest energy in the diagram is 0 eV. How can the energy be negative?
 
  • #5
DrClaude said:
The lowest energy in the diagram is 0 eV. How can the energy be negative?
ah i was doing E-U. would it be .6 ev? I've tried that though and it was wrong. I was doing this thinking that the particle would be found in region 2. after rereading the chapter i read that we can never observe the particle in the forbidden region. so does this mean that the particle will be found in region 3?
 
  • #6
gv3 said:
ah i was doing E-U. would it be .6 ev? I've tried that though and it was wrong. I was doing this thinking that the particle would be found in region 2. after rereading the chapter i read that we can never observe the particle in the forbidden region. so does this mean that the particle will be found in region 3?
i got the answer keeping the above in mind. it was 3 ev. I don't get conceptually though why the energy of the particle has to equal the potential in the third well in order for the particle to get stuck in the first well.
 
  • #7
gv3 said:
i got the answer keeping the above in mind. it was 3 ev. I don't get conceptually though why the energy of the particle has to equal the potential in the third well in order for the particle to get stuck in the first well.
If you find the particle in region III, it has a minimum energy of 3 eV (plus eventual kinetic energy). Likewise, it you find it in region II, it has at least 5 eV. Therefore, if the particle has an energy < 3 eV, it can only be found in region I, hence it is trapped there.
 
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  • #8
DrClaude said:
If you find the particle in region III, it has a minimum energy of 3 eV (plus eventual kinetic energy). Likewise, it you find it in region II, it has at least 5 eV. Therefore, if the particle has an energy < 3 eV, it can only be found in region I, hence it is trapped there.
So shouldn't the question be what the maximum energy would be for it to be trapped in the first well? Since it can't have more than 3 eV to be stuck in the first well.
 
  • #9
The phrasing is a bit weird.
 
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  • #10
DrClaude said:
The phrasing is a bit weird.
Thanks for helping! :D
 
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