Calculate E Min: n=1, L=2a, ψ=2 - Help!

In summary: So the ground state energy is half of the given value, which is 1eVIn summary, the ground state energy of a particle confined in a one-dimensional potential box with impenetrable walls at x = ±a is half of the first excited state energy, which is given to be 2eV. This implies that the value of the quantity ##\dfrac{\pi^2\hbar^2}{2ma^2}## is equal to 2 × 1.6e-19 J and the ground state energy is 1eV.
  • #1
tanaygupta2000
208
14
Homework Statement
A particle is confined in a one dimensional potential box with impenetrable walls at x = ±a. Its energy eigenvalue is 2eV and corresponds to the eigenfunction of the first excited state. What is the lowest possible energy of the particle ?
Relevant Equations
Hψ = Eψ
E = (nπ hbar)^2/2mL^2
For calculating minimum E, I have n = 1, L = 2a & ψ = 2.
But I am not getting how to exactly find its value.
Please help me !
 
Physics news on Phys.org
  • #2
tanaygupta2000 said:
Homework Statement:: A particle is confined in a one dimensional potential box with impenetrable walls at x = ±a. Its energy eigenvalue is 2eV and corresponds to the eigenfunction of the first excited state. What is the lowest possible energy of the particle ?
Relevant Equations:: Hψ = Eψ
E = (nπ hbar)^2/2mL^2

For calculating minimum E, I have n = 1, L = 2a & ψ = 2.
But I am not getting how to exactly find its value.
Please help me !
What is ##\psi = 2## supposed to mean?
 
  • #3
tanaygupta2000 said:
Its energy eigenvalue is 2eV and corresponds to the eigenfunction of the first excited state.
What does this statement imply about the value of the quantity ##\dfrac{\pi^2\hbar^2}{2ma^2}##?
 
  • #4
kuruman said:
What does this statement imply about the value of the quantity ##\dfrac{\pi^2\hbar^2}{2ma^2}##?
The value of this quantity should be equal to 2 × 1.6e-19 J
 
  • #5
tanaygupta2000 said:
The value of this quantity should be equal to 2 × 1.6e-19 J
Why is that?
 
  • #6
PeroK said:
Why is that?
Since this is quantized energy and it is given that the value of energy eigenvalue is 2eV
 
  • #7
tanaygupta2000 said:
Since this is quantized energy and it is given that the value of energy eigenvalue is 2eV
Which eigenvalue is ##2eV##? ##n = 1## or ##n = 2##?
 
  • #8
PeroK said:
Which eigenvalue is ##2eV##? ##n = 1## or ##n = 2##?
First excited state, n=1
 
  • #9
tanaygupta2000 said:
First excited state, n=1
So the ground state is n = 0?
 
  • #10
tanaygupta2000 said:
First excited state, n=1
And which energy eigenstate are you trying to find?
 
  • #11
Okay now I think I got it. The given value is for n=2 and I need to find it for n=1. Right ?
 
  • #12
tanaygupta2000 said:
Okay now I think I got it. The given value is for n=2 and I need to find it for n=1. Right ?
Right!
 
  • Like
Likes tanaygupta2000

1. What is the formula for calculating E Min?

The formula for calculating E Min is: E Min = n^2 * h^2 / 8 * m * L^2

2. What do the variables in the formula represent?

The variables in the formula represent the following:- n: principal quantum number- h: Planck's constant- m: mass of the particle- L: length of the potential well

3. What is the significance of n=1 in the formula?

The value of n=1 represents the ground state of the system. This is the lowest energy level that an electron can occupy in the potential well.

4. How is L=2a different from other values of L in the formula?

L=2a represents a specific length of the potential well, where a is the width of the well. This value is used when the potential well is symmetrical and the electron is confined to a specific region.

5. How can I use this formula to find the minimum energy for a specific system?

To use this formula, you will need to know the values of n, L, ψ, and the mass of the particle. Once you have these values, you can plug them into the formula and solve for E Min, which will give you the minimum energy for that particular system.

Similar threads

  • Introductory Physics Homework Help
Replies
28
Views
376
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
89
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
31
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
204
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
895
  • Introductory Physics Homework Help
Replies
3
Views
641
Back
Top