Number of standing waves in a potential barrier

AI Thread Summary
In the discussion about the number of standing waves in a potential barrier for a resonant tunneling diode, the effective mass of the electron is considered to be 0.9 times the free electron mass. The equation used to calculate the energy levels of standing wave states is E = (h^2 n^2)/(8mL^2). Initial calculations yielded an unreasonably high number of states, prompting a reevaluation. It was clarified that the value calculated represented n^2 for the last state fitting into the well, leading to the conclusion that the maximum number of states is actually 7 after taking the square root. This highlights the importance of careful mathematical handling in quantum mechanics problems.
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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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