Wavelength in finite potential well

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SUMMARY

The discussion centers on determining the wavelength (λ) of an electron in a finite square well potential, specifically with boundaries at ±400 pm. The wave function is expressed as A cos(2πx / λ), where A is a normalization constant. Participants estimated λ using various data points, ultimately concluding that the correct value is approximately 1950 pm, which falls within the required 10% accuracy. The conversation emphasizes the importance of using graphical estimations and iterative testing to refine answers in quantum mechanics problems.

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  • Understanding of quantum mechanics principles, particularly wave functions.
  • Familiarity with finite potential wells and their properties.
  • Knowledge of trigonometric functions and their applications in physics.
  • Ability to perform numerical estimations and iterative calculations.
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  • Study the properties of finite potential wells in quantum mechanics.
  • Learn about wave function normalization techniques.
  • Explore graphical methods for estimating solutions in quantum problems.
  • Investigate the implications of boundary conditions on wave functions.
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Students and educators in physics, particularly those focusing on quantum mechanics and wave phenomena, as well as anyone seeking to improve their problem-solving skills in this area.

Thefox14
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Homework Statement


1)The graph below represents the ground state wave function of an electron in a finite square well potential of width L. The potential is zero at x = 0.

FFX7a.gif


The wave function of the electron within the well is of the form A cos( 2πx / λ ) where A is a normalization constant.

What is the approximate value of λ, within about 10%?


Homework Equations


A cos( 2πx / λ ) and the left and right boundaries are at +- 400pm


The Attempt at a Solution



At 0pm cos(2pi x/lambda) = 1 so A has to be .075

then it looks like at 300pm y is .05 so:
.05 = .075*cos(2pi * 300pm / lambda)

lambda = 2pi * 300pm / cos-1(.075 / .05) = 2241pm but this is wrong
Or if i used 400pm as a data point: 2pi * 300pm / cos-1(.075 / .05) = 2041pm (still wrong)

What am i doing wrong here?
 
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Just helped a friend with a similar problem... so, I would agree with you that the left and right boundaries of the potential well appear to be around +/-400 pm. This is because of the change in the sign of the second derivative (which would be where the oscillating term, like sine or cosine turn into a decaying exponential). So, half of the wavelength must be (at least) 800 (to fill the width of the well). However, we know it's a bit larger than that since the function does not cross the x-axis by the time it hits the finite potential well wall. Basically, I think that the question is not looking for an analytically found answer but just using the graph to make estimations... extending the part of the function that looks like a wave, I would just say that half of lambda is around 1000pm. Try some things around 2000pm (your 2041pm seems pretty close, so I don't know what is wrong there: it should be within 10%). Overall, I hope my answer helps out a bit, but you may want to ask your teacher/professor about good estimation techniques for these kind of graphs. (My estimation for the final answer: 2000pm... from what you've already guessed, it might be a bit less than that)
 
Thats a great way of getting a really quick approximation to test your final answer for these things! Sadly I'm still not getting the right answer :(

I tried values from 1000-2500 (increments of 100), and still couldn't find what it was so I could "backtrack" and find what I was doing wrong.
 
well, the question says that it needs to be within 10%... I am 99% sure that the answer is greater than 1600pm (the boundaries of the well) and 90% it's around 2000pm... I would try values between 1800 and 2100 with smaller increments maybe? if it's an online submission form that you're using, it might be pickier than 10% (like: +/-5% or even less) so increments of 50 should give you the answer. If the answer isn't within this range, then I'm absolutely baffled... might even say that the question is incorrect. let me know if you figure out the answer, because now I'm really curious!
 
Wow...just wow. It wanted the answer of 1950 picometers haha

Thanks for the suggestion man!
 

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