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

AI Thread Summary
The discussion revolves around calculating the width of a finite square well given a ground-state energy of 2.50 eV. The initial approach incorrectly applied the infinite square well equation, leading to an incorrect width of 0.388 nm. Participants clarified that the problem involves a finite-depth well, requiring a different approach using the relationship U0=6E(1−IDW). After correcting the calculations and using the appropriate equations, the width was determined to be approximately 0.307 nm. Concerns were raised about the problem's clarity and relevance to the course material, prompting discussions about reporting it as poorly designed.
meaghan
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Homework Statement


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 ?

Homework Equations


En = h2n2/8mL2

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
 
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meaghan said:
U0=6E1−IDW.
I don't understand what that means.

meaghan said:

Homework Equations


En = h2n2/8mL2
That equation is only valid for an infinite square well.
 
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.
 
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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