# Approximate transmission coefficient of a square barrier

• bobby.pdx
In summary, the transmission coefficient for electrons of 7,000 meV is T(E)≈exp((-2/hbar)√(2m)∫√(U(x)-E)dx)
bobby.pdx

## Homework Statement

Two conductors are separated by an insulator. Model the insulator as a square barrier of height 0.01 keV and a width of 5nm. Determine the transmission coefficients for electrons of 7,000 meV.

The only thing is I have to use the approximation formula for finding the transmission coefficient of a barrier with arbitrary shape.

## Homework Equations

T(E)≈exp((-2/hbar)√(2m)∫√(U(x)-E)dx)

## The Attempt at a Solution

T(E)≈exp((-2/1.973keVA/c)√(2(511keV/c))∫(U(x)-E)dx)

I'm not sure what I'm supposed to do for the ∫(U(x)-E)dx part. I solved the problem using the formula for the transmission coefficient for a square barrier and I got T≈0.96x10^-38. I'm pretty certain this is correct because there was a very similar example in my book.

That formula had a (U-E) where I used 0.01keV-7000meV=.003keV

If I enter .003keV for U(x)-E then I would get ≈(.907)^x

And this is where I get stuck because I should somehow get an answer that is close to T≈0.96x10^-38
Any help would be appreciated!

Isn't U-E a constant, so you can bring it outside the integral ?

That's what I did. Still gave me the same answer

Can't see it being done. Not telepathic and not clearvoyant. Show what you do, so we can discuss it...

I showed you exactly what I did. The first calculation under "3. The Attempt at a Solution". I put 0.03keV for U-E and I entered it exactly like that into my calculator. I told you the answer I got and the answer I'm looking for. Not sure how I can be any more specific.

T(E)≈exp((-2/1.973keVA/c)√(2(511keV/c))∫(0.003keV)dx)=(0.907)^x

It should look like 0.96x10^-38

sorry I get exp(-.097x)

bobby.pdx said:
T(E)≈exp((-2/1.973keVA/c)√(2(511keV/c))∫(0.003keV)dx)=(0.907)^x

Did you drop a square root somewhere? Compare what you wrote here with your original expression in the "Relevant Equations" section.

Note, you wrote the units of the mass of the electron as kev/c. Is this correct?

I made a mistake in that calculation. The answer I got from that equation is actually exp(-.097x) which is still not correct. Also I meant to write the unit of mass as keV/c^2

I don't get .097. Again, did you drop a square root? I see a square root symbol occurring twice in

T(E)≈exp((-2/hbar)√(2m)∫√(U(x)-E)dx)

Oh you're completely right. Now I got (.169)^x. Now if I change 5nm to 50A I get (.169)^50=.287x10^-38. This is a way better answer even though it's not the same it's at least the same order of magnitude.

That's close to what I get, too. I get Exp[-1.77x] = Exp[-88.5] = .37x10-38

Last edited:
cool. thanks for the help

## 1. What is the approximate transmission coefficient of a square barrier?

The approximate transmission coefficient of a square barrier is the probability that a particle will pass through the barrier when it encounters it.

## 2. How is the transmission coefficient of a square barrier calculated?

The transmission coefficient is calculated using the Schrödinger equation, which takes into account the energy of the particle and the height and width of the barrier.

## 3. What factors affect the approximate transmission coefficient of a square barrier?

The approximate transmission coefficient of a square barrier is affected by the energy of the particle, the height and width of the barrier, and the shape and material of the barrier.

## 4. Can the transmission coefficient of a square barrier be greater than 1?

No, the transmission coefficient of a square barrier cannot be greater than 1 as it represents a probability and probabilities cannot exceed 1.

## 5. How does the transmission coefficient of a square barrier change with increasing barrier height?

As the barrier height increases, the transmission coefficient decreases. This is because a higher barrier means a higher energy barrier for the particle to pass through, making it less likely to pass through the barrier.

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