Potential Barrier (quantum mechanics)

I have one more question. Is 1.675x10^-27 the mass in kg of one neutron?Yes, that is the mass of a neutron in kilograms.
  • #1
roam
1,271
12

Homework Statement



I need help with part (iii) and (ii) of the follwoing problem:

http://img228.imageshack.us/img228/516/79620850.jpg

Homework Equations



From my notes transmission coefficient for E<U is:

[tex]T(E)= \left[ 1+\frac{1}{4} \left( \frac{U^2}{E(U-E)} \right) sinh^2 (\alpha L) \right]^{-1}[/tex]

And for E>U it is:

[tex]T(E)= \left[ 1+\frac{1}{4} \left( \frac{U^2}{E(E-U)} \right) sin^2 (k' L) \right]^{-1}[/tex]

Where [tex]k'=\frac{\sqrt{2m(E-U)}}{\hbar}[/tex] and [tex]\alpha =\frac{\sqrt{2m(U-E)}}{\hbar}[/tex].

The Attempt at a Solution



So for part (ii). I used the first equation and got a value of 4.59x10-4. This doesn't seem like a correct number for the number of neutrons. So do I have to multiply this with the total number of neutrons? :confused:

For (iii) I am confused and I don't know which equation to use. We know the equation when E<U and when E>U, but what exactly is the equation for when E=U?

I don't quite understand the hint, does this mean we need the following equation:

[tex]T(E)= \left[ 1+\frac{1}{4} \left( \frac{U^2}{E(U-E)} \right) \alpha^2 L^2 \right]^{-1}[/tex]

Is that right?
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
roam said:

Homework Statement



I need help with part (iii) and (ii) of the follwoing problem:

http://img228.imageshack.us/img228/516/79620850.jpg

Homework Equations



From my notes transmission coefficient for E<U is:

[tex]T(E)= \left[ 1+\frac{1}{4} \left( \frac{U^2}{E(U-E)} \right) sinh^2 (\alpha L) \right]^{-1}[/tex]

And for E>U it is:

[tex]T(E)= \left[ 1+\frac{1}{4} \left( \frac{U^2}{E(E-U)} \right) sin^2 (k' L) \right]^{-1}[/tex]

Where [tex]k'=\frac{\sqrt{2m(E-U)}}{\hbar}[/tex] and [tex]\alpha =\frac{\sqrt{2m(U-E)}}{\hbar}[/tex].

The Attempt at a Solution



So for part (ii). I used the first equation and got a value of 4.59x10-4. This doesn't seem like a correct number for the number of neutrons. So do I have to multiply this with the total number of neutrons? :confused:
What does T(E) represent? If you understand that, it should be clear how to proceed.

For (iii) I am confused and I don't know which equation to use. We know the equation when E<U and when E>U, but what exactly is the equation for when E=U?

I don't quite understand the hint, does this mean we need the following equation:

[tex]T(E)= \left[ 1+\frac{1}{4} \left( \frac{U^2}{E(U-E)} \right) \alpha^2 L^2 \right]^{-1}[/tex]

Is that right?
Yes. Substitute in for ##\alpha## now.
 
Last edited by a moderator:
  • #3
vela said:
What does T(E) represent? If you understand that, it should be clear how to proceed.

Thank you for your reply. I believe T(E) represents the probability of the particles being transmitted relative to the total incident particles. So isn't it correct then to multiply it by the total number of particles to see how many are transmitted?

In that case I will have (106)x(4.59x10-4)=459

Yes. Substitute in for ##\alpha## now.

But doesn't ##\alpha## equal to 0 when U=E?

##\alpha = \frac{\sqrt{2(1.675\times 10^{-27})(4MeV-4MeV)}}{1.054 \times 10^{-34}} = 0##

This will make the whole T(E) expression equal to 1, so that means that all the particles get transmitted? Is that right or did I make a mistake?
 
  • #4
roam said:
Thank you for your reply. I believe T(E) represents the probability of the particles being transmitted relative to the total incident particles. So is isn't it correct then to multiply it by the total number of particles to see how many are transmitted?

In that case I will have (106)x(4.59x10-4)=459
Exactly, T(E) is the fraction that gets transmitted.



But doesn't ##\alpha## equal to 0 when U=E?

##\alpha = \frac{\sqrt{2(1.675\times 10^{-27})(4MeV-4MeV)}}{1.054 \times 10^{-34}} = 0##

This will make the whole expression equal to 1, so that means that all the particles get transmitted? Is that right or did I make a mistake?
No, remember the expression for T(E) is valid for E<U, so U-E isn't 0. Do the substitution and simplify, and then take the limit as E approaches U.
 
  • #5
vela said:
Exactly, T(E) is the fraction that gets transmitted.

Thanks, so I guess it is correct to multiply T(E) by the total number of particles.

No, remember the expression for T(E) is valid for E<U, so U-E isn't 0. Do the substitution and simplify, and then take the limit as E approaches U.

I see. So here is what I did:

##\lim_{E \to U} \left( 1+ \frac{1}{4}\frac{U^2}{E(U-E)} . \frac{2m(U-E)}{\hbar^2} L^2 \right)^{-1}##

##\lim_{E \to U} \left( 1+ \frac{mU^2}{2 \hbar E} L^2 \right)^{-1}##

And since for small x we have (1+x)-1=1-x,

##\lim_{E \to U} \left( 1- \frac{mU^2}{2 \hbar E} L^2 \right)^{-1} = \frac{mU^2}{2 \hbar U} L^2##

Is this correct now?
 
  • #6
No, you made algebra mistakes, and you shouldn't approximate using a series.
 
  • #7
Oops that was a typo, here is what I've got

[tex]\lim_{E \to U} \left( 1+ \frac{mU^2}{2 \hbar^2 E} L^2 \right)^{-1}[/tex]

I have substituted the limits first and it is possible to take the inverse later:

##T(x)=\left( 1+ \frac{mU}{2 \hbar^2} L^2 \right)^{-1}##

Is that fine now?
 
  • #8
Looks fine.
 
  • #9
Alright, thank you so much for the help.
 

Related to Potential Barrier (quantum mechanics)

1. What is a potential barrier in quantum mechanics?

A potential barrier in quantum mechanics refers to a region in space where the potential energy is higher than the energy of a particle. This barrier can prevent the particle from moving through the region, creating a barrier for its motion.

2. How does a potential barrier affect the behavior of a particle?

A potential barrier can cause a particle to behave differently depending on its energy. If the particle has enough energy, it can tunnel through the barrier and continue its motion. If the particle does not have enough energy, it will be reflected by the barrier and its motion will be altered.

3. What is the difference between a finite and an infinite potential barrier?

A finite potential barrier has a finite width and height, meaning that there is a chance for the particle to tunnel through it. An infinite potential barrier, on the other hand, has an infinitely high potential energy, making it impossible for the particle to tunnel through it.

4. How is the transmission coefficient used to describe a potential barrier?

The transmission coefficient is a measure of the probability that a particle will successfully tunnel through a potential barrier. It is used to describe the likelihood of the particle's motion being transmitted through the barrier, rather than being reflected by it.

5. What real-life applications does the concept of potential barriers have?

Potential barriers are important in understanding the behavior of electrons in semiconductors, which are used in electronic devices such as transistors and diodes. They are also relevant in studying nuclear reactions and the stability of atomic nuclei.

Similar threads

  • Advanced Physics Homework Help
Replies
0
Views
546
  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
429
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
24
Views
1K
  • Advanced Physics Homework Help
Replies
10
Views
821
  • Advanced Physics Homework Help
Replies
5
Views
1K
  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
8
Views
1K
  • Advanced Physics Homework Help
Replies
3
Views
984
Back
Top