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

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To calculate the minimum energy (E) for a particle in a one-dimensional potential box, the relevant parameters are n = 1, L = 2a, and an energy eigenvalue of 2 eV corresponding to the first excited state (n = 2). The equation E = (nπ hbar)^2 / 2mL^2 is used to find the energy, with the understanding that the eigenvalue of 2 eV indicates quantized energy levels. The discussion clarifies that the ground state corresponds to n = 0, while the first excited state is n = 1. Participants confirm that the given energy value is for n = 2, and the goal is to determine the energy for n = 1. The calculations are essential for understanding the energy states of the particle in the potential box.
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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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