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Old Nov 14th 2014, 02:19 PM   #1
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particle (infinite 1-D well)

The problem statement, all variables and given/known data
If a particle (infinite 1-D well) in ground state n =1 with an energy 1.26 eV above E=0. Whats the energy needed to get it to 3rd excited state n =4?

any hints?
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Old Nov 14th 2014, 03:00 PM   #2
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Originally Posted by mss90 View Post
The problem statement, all variables and given/known data
If a particle (infinite 1-D well) in ground state n =1 with an energy 1.26 eV above E=0. Whats the energy needed to get it to 3rd excited state n =4?

any hints?
The energy of a particle in an infinite square well potential (V = 0 from x = 0 to x = a) is
E = [pi ^2 hbar^2 / (2 m a^2)] n^2

You can use this for n = 1, E = 1.26 eV to find a.

-Dan
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Old Nov 15th 2014, 08:47 AM   #3
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whats the name of that formula? thanks
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Old Nov 15th 2014, 01:48 PM   #4
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Did you mean En = h2kn2/(2m) or En = n2π2ħ2/(2mL2)?

I got


a=(pi^2 ħ^2/E2m)*1^2 where ħ=h/2pi
a=1.66E-34/2.3E-30
a= 7.22E-5eV

Last edited by mss90; Nov 15th 2014 at 02:03 PM.
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Old Nov 15th 2014, 02:05 PM   #5
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Originally Posted by mss90 View Post
Did you mean En = h2kn2/(2m) or En = n2π2ħ2/(2mL2)?

Whats ħ anyways?
The second expression is correct. I used "a" for the width of the square well, whereas you used "L."

The Schrodinger equation is a "eigenvalue" equation. E is the energy eigenvalue for a particle in a state with "n" equal to some (quantized) number.

How can you not know what hbar is?? hbar is defined as Planck's constant divided by 2 pi.

How can you even ask these questions? You had to solve the Schrodinger equation to get to this point!

-Dan
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Old Nov 15th 2014, 02:10 PM   #6
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Why should I know that, I havnt learned it anywhere.. I removed it from the question as i managed to find it online. No I havnt solved schrodinger equation, and I found the formula online.
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Old Nov 15th 2014, 02:16 PM   #7
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so if i just got the width of the well, how do i go about finding the energy needed to get it to 3rd excited state n =4
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Old Nov 15th 2014, 04:18 PM   #8
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Originally Posted by mss90 View Post
so if i just got the width of the well, how do i go about finding the energy needed to get it to 3rd excited state n =4
Just plug n = 4 into the energy equation I posted.

I'm curious. How can you be expected to solve the problem of an infinite square well if you haven't covered the Schrodinger equation? This makes no sense to me at all.

-Dan
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Old Nov 16th 2014, 06:24 AM   #9
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like this -> E4=(1.66E-34/(2m*7.22E-52))*42
E4=1.6E27eV?

well it has been covered briefly but i didnt learn it. To be honest I never expected there to be this much maths and physics in the degree im doing. I havnt done physics and maths in almost 4 years now and no previous knowledge in them were required. Hence Im not sure If ill manage. I appreciate you assisting me tho
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Old Nov 16th 2014, 10:53 AM   #10
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You have an energy level of 1.5 x 10^{27} eV?? Ouch! Any electron orbital energies will be in the eV range, not 10^{27} eV.

Two things first: Notation. "42" is a number. It is not 16. When writing exponents please use "^" ie 4^2.

Second: I can help you but I can't really teach you on the Forum. You really need to talk to your instructor about these things.

Okay. We don't yet know m or L. So I'm going to go for broke and solve
E_n = (hbar^2 Pi^2) n^2/ (2mL^2)

for 2mL^2: 2mL^2 = (hbar^2 Pi^2) n^2 / E_n

We are given n =1, E_1 = 1.26 eV, which gives:
2mL^2 = 3.39348 x 10^{-30}

Now we can use that for n = 4:
E_n = (hbar^2 Pi^2) n^2/ (2mL^2)

E_4 = (hbar^2 Pi^2 * 4^2) / 3.39348 x 10^{-30} = 20.16 eV.

If you do a few of these you will notice that E_n = E_1 * n^2.

-Dan
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Last edited by topsquark; Nov 16th 2014 at 10:55 AM.
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