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 Nuclear and Particle Physics Nuclear and Particle Physics Help Forum Nov 14th 2014, 02:19 PM #1 Junior Member   Join Date: Oct 2014 Posts: 12 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?   Nov 14th 2014, 03:00 PM   #2
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 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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See the forum rules here.   Nov 15th 2014, 08:47 AM #3 Junior Member   Join Date: Oct 2014 Posts: 12 whats the name of that formula? thanks   Nov 15th 2014, 01:48 PM #4 Junior Member   Join Date: Oct 2014 Posts: 12 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.   Nov 15th 2014, 02:05 PM   #5
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 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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See the forum rules here.   Nov 15th 2014, 02:10 PM #6 Junior Member   Join Date: Oct 2014 Posts: 12 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.   Nov 15th 2014, 02:16 PM #7 Junior Member   Join Date: Oct 2014 Posts: 12 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   Nov 15th 2014, 04:18 PM   #8
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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
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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See the forum rules here.   Nov 16th 2014, 06:24 AM #9 Junior Member   Join Date: Oct 2014 Posts: 12 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   Nov 16th 2014, 10:53 AM #10 Forum Admin   Join Date: Apr 2008 Location: On the dance floor, baby! Posts: 2,780 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 __________________ Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup. See the forum rules here. Last edited by topsquark; Nov 16th 2014 at 10:55 AM.  Tags infinite, particle Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post roger Quantum Physics 0 Jan 31st 2015 10:14 AM eberth_quezada Quantum Physics 3 Oct 9th 2011 03:44 AM BloodBaneZXY Electricity and Magnetism 2 May 3rd 2010 01:12 AM jonbrutal Atomic and Solid State Physics 1 Feb 12th 2010 10:43 AM synclastica_86 Quantum Physics 1 Oct 1st 2009 09:00 AM 