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Infinite well width?

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  1. Apr 25, 2016 #1
    1. The problem statement, all variables and given/known data
    An electron is bound in a square well of depthU0=6E1−IDW.

    What is the width of the well if its ground-state energy is 2.50 eV ?

    2. Relevant equations
    En = h2n2/8mL2
    3. The attempt at a solution
    I used n = 1
    so I get:
    25eV*1.6*10-19 = h2/8*9.11*10-31*L2

    I got L = .388 nm. It wasn't correct. I ignored the depth part at the beginning since I thought it was irrelevant, but it might not be
     
  2. jcsd
  3. Apr 25, 2016 #2

    DrClaude

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    Staff: Mentor

    I don't understand what that means.

    That equation is only valid for an infinite square well.
     
  4. Apr 27, 2016 #3
    Oh man, this was one of my homework problems (assigned on Monday) too.

    You've got [U0=6E(1−IDW)]. "IDW" I'm pretty sure means Infinite-Depth Well. This is a finite-depth well so you cannot use infinite-depth equations, but this provided equation gives you a way to apply IDW equations to the FDW.

    In practical terms - what they should have included in the problem - is that we can treat [U0=6E(1−IDW)] as [U0=6E].

    This is where I don't know where to go. So I sacrificed my own problem to the "give up" button, took my own answer, divided it by my U0, multiplied it by yours and then put it into the IDW equation. What I got was L = 0.3064 nm. Maybe that will work.

    I still have no idea why they thought this was a good problem. I will be writing complaints and talking to my professor.
     
  5. Apr 28, 2016 #4
    Belay that. I asked my professor and this is the real answer:

    U0 is the depth of the finite well; E(1−IDW) is the depth of the infinite well (roll with it). With this equation you find that [E1 = 0.625E(1−IDW)] - how you're supposed to figure this out I'm not sure, but you have to to solve the problem.

    E1=0.625E(1−IDW)=2.5eV

    E(1−IDW)=(2.5eV)/0.625=h2n2/8mL2

    with n=1 and m=9.11*10-31kg, and remembering to convert eV to J

    L=0.306596nm (with sigfigs, 0.307)

    I used this solution to confirm with my numbers and it worked out. I'm still going to report it as a bad problem, though, because at least in my course we didn't cover this in lecture and the textbook *does* have it but in a very roundabout way that a lot of people probably wouldn't catch.
     
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