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

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Homework Help Overview

The discussion revolves around calculating the minimum energy of a particle confined in a one-dimensional potential box, with specific parameters including n = 1 and L = 2a. The energy eigenvalue is given as 2eV, and there is some confusion regarding the implications of the eigenfunction and the states involved.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to clarify the meaning of the eigenfunction value and its relation to the energy eigenvalue. Questions arise about the implications of the given energy eigenvalue and which quantum state it corresponds to.

Discussion Status

The discussion is active, with participants exploring the relationship between the energy eigenvalue and the quantum states. Some have identified that the given energy corresponds to n=2, while others are trying to confirm the understanding of the ground state and first excited state definitions.

Contextual Notes

There is a lack of clarity regarding the definition of the eigenfunction and how it relates to the energy levels. Participants are also navigating the implications of quantized energy in the context of the problem.

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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 !
 
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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?
 
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}##?
 
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
 
tanaygupta2000 said:
The value of this quantity should be equal to 2 × 1.6e-19 J
Why is that?
 
PeroK said:
Why is that?
Since this is quantized energy and it is given that the value of energy eigenvalue is 2eV
 
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##?
 
PeroK said:
Which eigenvalue is ##2eV##? ##n = 1## or ##n = 2##?
First excited state, n=1
 
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!
 
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