Number of standing waves in a potential barrier

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SUMMARY

The discussion centers on calculating the number of standing wave states in a resonant tunneling diode structure with two AlAs barriers of height 1.2 eV and width 2.4 nm, enclosing a well of width 4.4 nm. Using the equation E = (h² n²) / (8mL²), the effective mass of the electron is taken as 0.9 times the free electron mass. The calculated energy for the lowest standing wave state is 0.0216 eV, leading to a maximum of 7 standing wave states fitting within the well, correcting an initial miscalculation that suggested 55 states.

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  • Understanding of quantum mechanics principles related to standing waves
  • Familiarity with resonant tunneling diodes and their structures
  • Knowledge of the effective mass of electrons in semiconductor materials
  • Proficiency in using the energy quantization formula E = (h² n²) / (8mL²)
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Homework Statement


Consider a resonant tunneling diode structure (attached image). This shows 2 AIAs barriers of height 1.2 eV and width t = 2.4 nm, enclosing a well of width L = 4.4 nm.

If the effective mass of the electron is taken as 0.9 times the free electron mass how many separate standing wave states n = 1, 2 ... do you think will be formed (Find th energies of the lwest standing wave sttes, assuming the barriers are infinitely tall, and compare the energy with the actual barrier height, 1.2 eV).

Homework Equations


<br /> E = \frac{h^2 n^2}{8mL^2}<br />


The Attempt at a Solution


I plugged the numbers into the equation above, but the number of wave states I got didn't make sense. It was way too high. The reason I think it is too high is because the question after this asks to find the probability of tunneling for each of these wave states that I find, and 55 is an unreasonable number of states.

<br /> E = \frac{(6.626 \times 10^{-34} ~Js)^2(1)^2}{8 \times 0.9 \times 9.1 \times 10^{-31} ~ kg \times (4.4 nm)^2} \\<br /> E = 3.46 \times 10^{-21} J = 0.0216 ~eV \\<br /> \frac{E_t}{E} = 1.2 ~ eV \div 0.0216 ~eV = 55.5 <br />
 
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That would be the case if the energy levels would be equally spaced.
What you found is the value of n^2 for the last state that will "fit" into the well.
So n of the last state is 7.
 
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nasu said:
That would be the case if the energy levels would be equally spaced.
What you found is the value of n^2 for the last state that will "fir" into the well.
So n of the last state is 7.

Oops, forgot to take the square root didn't I? Thanks!
 

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