Physics Help Forum First 6 allowed states for an electron in an infinite square well of width 0.2nm

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 Mar 23rd 2009, 03:00 AM #1 Junior Member   Join Date: Jul 2008 Location: sydney Posts: 14 First 6 allowed states for an electron in an infinite square well of width 0.2nm Hi, just want to check if this is correct. Question asks to find the first 6 allowed states for an electron in infinite square well of width 0.2nm (Partial answer E1=9.4eV E2=37.6eV) For an infinite square well En is given by $\displaystyle E_n = \frac{{n^2 \pi ^2 \hbar ^2 }}{{2mL^2 }}$ I just went that E being proportional to n^2 than E at n=3 would be 3^2 time E3 thus giving 84.6eV and similarly E4=150.4eV , E5=235eV and E6=338.4eV Or should have this been done differently?
 Mar 23rd 2009, 03:40 AM #2 Physics Team   Join Date: Feb 2009 Posts: 1,425 That is perfectly fine except you haven’t mentioned E2 = 37.6 ev.
 Mar 23rd 2009, 04:00 AM #3 Junior Member   Join Date: Jul 2008 Location: sydney Posts: 14 Thanks. I didn't mention E2 as it was E1 and E2 were already given.

 Tags 02nm, allowed, electron, infinite, square, states, width

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