# Particle in a box

## 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.

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PeroK
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Gold Member
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.
What is ##\psi = 2## supposed to mean?

kuruman
Homework Helper
Gold Member
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}##?

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

PeroK
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The value of this quantity should be equal to 2 × 1.6e-19 J
Why is that?

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

PeroK
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Gold Member
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##?

Which eigenvalue is ##2eV##? ##n = 1## or ##n = 2##?
First excited state, n=1

kuruman
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First excited state, n=1
So the ground state is n = 0?

PeroK
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Gold Member
First excited state, n=1
And which energy eigenstate are you trying to find?

Okay now I think I got it. The given value is for n=2 and I need to find it for n=1. Right ?

PeroK
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