Particle in a box

  • #1
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 !
 

Answers and Replies

  • #2
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
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
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
 
  • #6
Why is that?
Since this is quantized energy and it is given that the value of energy eigenvalue is 2eV
 
  • #7
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##?
 
  • #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
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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