Finite potential well problem penetration depth

In summary, the autograder says that my answer of 0.434nm is incorrect because I have substituted values given E and Vo and mass of electron and hket in the above formula for penetration depth in case of finite potential well. The formula is independent of probability or probability amplitude, and the autograder says that my answer of 0.434nm is incorrect. I have just substituted values given E and Vo and mass of electron and hket in the above formula for penetration depth in case of finite potential well. What has probability and probability amplitude got to do with the question?? You can yourself check and using the formula you will get value of 0.434nm.
  • #1
Ashish Somwanshi
31
4
Homework Statement
Consider a potential step with a height of V0=0.5eV. An electron is incident from the lower potential side (where the potential energy is set to equal to zero) with an energy E=0.3eV.

Find the penetration depth which is defined as the distance into the barrier region at which the probability decreases to 1/e of its initial value at the potential discontinuity, z=0.

Give your answer in unit of nm. Answers within 5% error will be considered correct.
Relevant Equations
Eta = hket/(2m(Vo-E)), where eta is the penetration depth and hket=1.05×10^(-34)
I don't understand where I went wrong, the formula and calculations which I have attached are correct...please do help if anyone can spot the mistake.
 

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  • #2
If possible, next time please type in your solution rather than post pictures of handwritten solutions.

Regarding your solution, did you take into account the distinction between "probability" and "probability amplitude"?
 
  • #3
Formula is independent of probability or probability amplitude and autograder says my answer of 0.434nm is incorrect. I have just substituted values given E and Vo and mass of electron and hket in the above formula for penetration depth in case of finite potential well. What has probability and probability amplitude got to do with the question?? You can yourself check and using the formula you will get value of 0.434nm.
 

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  • #4
Ashish Somwanshi said:
Formula is independent of probability or probability amplitude and autograder says my answer of 0.434nm is incorrect. I have just substituted values given E and Vo and mass of electron and hket in the above formula for penetration depth in case of finite potential well. What has probability and probability amplitude got to do with the question?? You can yourself check and using the formula you will get value of 0.434nm.
It is indeed true that if one uses the formula that you provided, one will get the answer you got. This means that you substituted the numbers in the formula correctly. However, what makes you think that the formula $$\eta =\frac{\hbar}{\sqrt{2m(V_0-E)}}$$ gives the length at which the probability drops to ##1/e## of its value at ##z=0##? In your notes you say "It can be derived". Fair enough. Please provide your derivation because I don't believe that the formula is independent from the probability or probability amplitude. The two drop to ##1/e## of their initial values at different distances from the discontinuity. I think you calculated the wrong ##\eta## but without its derivation I cannot be sure. However, it seems that the autograder agrees with me.
 
  • #5
Ashish Somwanshi said:
Formula is independent of probability or probability amplitude and autograder says my answer of 0.434nm is incorrect. I have just substituted values given E and Vo and mass of electron and hket in the above formula for penetration depth in case of finite potential well. What has probability and probability amplitude got to do with the question?? You can yourself check and using the formula you will get value of 0.434nm.

You have two different definitions of penetration depth.

First definition:

1665168909246.png


Second definition.
1665168979209.png


The first definition is the distance where the probability is reduced by 1/e. The second definition is the distance where the wave function (also known as the probability amplitude) is reduced by 1/e. Your answer is found using the second definition. Maybe they want the answer based on the first definition.
 
  • #6
TSny said:
The first definition is the distance where the probability is reduced by 1/e. The second definition is the distance where the wave function (also known as the probability amplitude) is reduced by 1/e. Your answer is found using the second definition. Maybe they want the answer based on the first definition.
Yes, the formula used by OP is for the wavefunction, not the probability.
 
  • #7
kuruman said:
It is indeed true that if one uses the formula that you provided, one will get the answer you got. This means that you substituted the numbers in the formula correctly. However, what makes you think that the formula $$\eta =\frac{\hbar}{\sqrt{2m(V_0-E)}}$$ gives the length at which the probability drops to ##1/e## of its value at ##z=0##? In your notes you say "It can be derived". Fair enough. Please provide your derivation because I don't believe that the formula is independent from the probability or probability amplitude. The two drop to ##1/e## of their initial values at different distances from the discontinuity. I think you calculated the wrong ##\eta## but without its derivation I cannot be sure. However, it seems that the autograder agrees with me.
There was no derivation of this formula in my notes. I found this formula from the net. It seems now I have to provide my original notes for further inspection. As soon as I finish writing my complete notes. I will post it here.
 
  • #8
Hi @Ashish Somwanshi. Your notes say that ##\eta## is the distance for the wavefunction (##\psi(z)##) to fall from ##\psi (L)## to ##\frac 1e \psi(L)##. That’s where your equation ##\psi(L+\eta)## = ##\frac 1e \psi(L)## comes from.

But the question is asking you to find “the distance into the barrier region at which the probability decreases to 1/e of its value at z=0”.

This distance is not ##\eta##, but is closely related to it.

##\psi(z)## does not give probability. Do you know how to find probability (or more correctly, probability density) from ##\psi##? It’s one of the most basic things you first learn about wave functions.
 
  • #9
Just square the absolute value of wavefunction it gives the probability density. So how do I find the penetration depth where the probability density reduces to 1/eof its initial value at z=0. There must be some well knownformula? Is it related to transmission probability?
 
  • #10
Inside the step, the wavefunction behaves as ##\psi(z) = A e^{- z/\eta}## where ##A## is a constant and ##z## is the penetration distance into the step. [Edit: Here, ##\eta = \frac{\hbar}{\sqrt{2m(V_0-E))}}##.]

How does ##|\psi(z)|^2## behave?
 
Last edited:
  • #11
TSny said:
Inside the step, the wavefunction behaves as ##\psi(z) = A e^{- z/\eta}## where ##A## is a constant and ##z## is the penetration distance into the step.

How does ##|\psi(z)|^2## behave?
Psi^2(z)=A^2*e^(-2z/eta)
 
  • #12
Ashish Somwanshi said:
Psi^2(z)=A^2*e^(-2z/eta)
Yes. Can you work with this expression to find the answer?
 
  • #13
Ashish Somwanshi said:
There must be some well knownformula? Is it related to transmission probability?
There is no well known formula. You have to derive it yourself. Just think. What is the probability at ##z=0## and at ##z=\eta##? Then write an equation that says that the latter is equal to ##1/e## of the former and solve for ##\eta##.
 
  • #14
@Ashish Somwanshi : In post #11 I was assuming that ##\eta## is the quantity ## \frac{\hbar}{\sqrt{2m(V_0-E))}}##, as given in your handwritten notes. Thus, at ##z = \eta## the wavefunction has decayed to ##\large \frac{1}{e}## of its value at ##z = 0##.
 
  • #15
I got A^2=(1/e)A^2, now how I am supposed to solve for eta?
 
Last edited:
  • #16
I am not getting the value of eta from method suggested by @kuruman
 
  • #17
Isn't this problem already solved at chegg? Can anyone can check it up on chegg and give a hint or method of action?
 
  • #18
If ##\eta = \frac{\hbar}{\sqrt{2m(V_0-E))}}##, then ##\psi(z) = A e^{- z/\eta}## and ##\psi^2(z) = A^2 e^{-2 z/\eta}##.

So, what is ##\psi^2(z)## when ##z = 0##? For what value of ##z## will ##\psi^2(z)## be reduced to 1/e of the value at ##z = 0##?
 
  • #19
TSny said:
If ##\eta = \frac{\hbar}{\sqrt{2m(V_0-E))}}##, then ##\psi(z) = A e^{- z/\eta}## and ##\psi^2(z) = A^2 e^{-2 z/\eta}##.

So, what is ##\psi^2(z)## when ##z = 0##? For what value of ##z## will ##\psi^2(z)## be reduced to 1/e of the value at ##z = 0##?
Thanks extremely, @kuruman and @TSny , I've passed the course and got the correct answer of 0.217nm for penetration depth for which the probability density reduces to 1/2.71828 of its initial value at z=0 of the potential step.
 
  • #20
Ashish Somwanshi said:
I am not getting the value of eta from method suggested by @kuruman
Please stop telling us you got wrong answer and start showing us what you did and how.
Ashish Somwanshi said:
Isn't this problem already solved at chegg? Can anyone can check it up on chegg and give a hint or method of action?
Chegg is for losers and Physics Forums is for learners. Which would you rather be?
Ashish Somwanshi said:
Thanks extremely, @kuruman and @TSny , I've passed the course and got the correct answer of 0.217nm for penetration depth for which the probability density reduces to 1/2.71828 of its initial value at z=0 of the potential step.
You are welcome and congratulations. Now sit back and contemplate what you learned from this entire experience about yourself and how you approach your studies. You don't have to tell us.
 

Related to Finite potential well problem penetration depth

1. What is a finite potential well problem?

A finite potential well problem is a concept in quantum mechanics where a particle is confined to a certain region by a potential barrier. This can be represented mathematically as a finite potential well, which is a potential energy function that has a finite value within a certain range and is infinite outside of that range.

2. What is the penetration depth in a finite potential well problem?

The penetration depth in a finite potential well problem refers to the distance that a particle can travel into the potential barrier before being reflected back. It is a measure of the particle's ability to penetrate through the barrier and is dependent on the energy of the particle and the height and width of the potential well.

3. How is the penetration depth calculated in a finite potential well problem?

The penetration depth can be calculated using the Schrödinger equation, which describes the wave function of a particle in a potential well. The wave function can be solved for both inside and outside the well, and the penetration depth is determined by the point where the two solutions match.

4. What factors affect the penetration depth in a finite potential well problem?

The penetration depth is affected by the energy of the particle, the height and width of the potential well, and the shape of the potential barrier. Higher energy particles and wider, taller barriers will have a larger penetration depth, while lower energy particles and narrower, shorter barriers will have a smaller penetration depth.

5. What is the significance of the penetration depth in a finite potential well problem?

The penetration depth is an important concept in quantum mechanics as it allows us to understand the behavior of particles in confined spaces. It also has practical applications in fields such as semiconductor physics, where the penetration depth of electrons determines the conductivity of a material. Additionally, the penetration depth can be used to study the properties of potential barriers and the behavior of particles in different energy states.

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