Quantum mechanics - several constant potentials

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  • #1
Eitan Levy
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Homework Statement:
A particle with mass m is in a one dimensional potential, as seen below.

The wave function in [itex]x<0[/itex] is 0.

The wave function in [itex]0<x<b[/itex] is: [itex]Asin(kx)[/itex]

The wave function in [itex]x>b[/itex] is: [itex]Be^{-\alpha x}[/itex]

It is known that [itex]k=3*10^{10}[/itex] and [itex]b=0.5333333*10^{-10}[/itex]

Find [itex]\alpha[/itex]
Relevant Equations:
Schrodinger stationary equation
Capture.PNG


What I tried to do was using the fact that the wave function should be continuous.

[itex]Asin(kb)=Be^{-\alpha b}[/itex]

The derivative also should be continuous:

[itex]kAcos(kb)=-\alpha Be^{-\alpha b}[/itex]

And the probability to find the particle in total should be 1:

[itex]\int_0^b A^2sin^2(kx) dx + \int_b^{\infty} B^2e^{-2\alpha x} dx =1 [/itex]

This set of equations is to hard to deal with, the equations should be solved with calculator only so I think I did something wrong.

Also, there may be a better way to approach this problem, but I'm not seeing it.
 

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  • #2
Steve4Physics
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[itex]Asin(kb)=Be^{-\alpha b}[/itex]

[itex]kAcos(kb)=-\alpha Be^{-\alpha b}[/itex]

[itex]\int_0^b A^2sin^2(kx) dx + \int_b^{\infty} B^2e^{-2\alpha x} dx =1 [/itex]
Hints: It's much easier than you think. You only need the first 2 equations.
 
  • #3
Eitan Levy
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Hints: It's much easier than you think. You only need the first 2 equations.
Is the answer 0.08763*10^(10)?
 
  • #4
Steve4Physics
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Is the answer 0.08763*10^(10)?
I haven't done the calculation. How did you get that value?
Also, have you forgotten units or were there really none supplied in the original question?
 
  • #5
Eitan Levy
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I haven't done the calculation. How did you get that value?
Also, have you forgotten units or were there really none supplied in the original question?
I divided the first equation by the second:

[itex]\frac{1}{k}tan(kb)=-\frac{1}{\alpha}[/itex]

And I plugged in the numbers.

No units were provided.
 
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  • #6
Steve4Physics
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I divided the first equation by the second:

[itex]\frac{1}{k}tan(kb)=-\frac{1}{\alpha}[/itex]

And I plugged in the numbers.

No units were provided.
Agreed. Well done! A couple of points:
1) You haven't correctly used standard form (mantissa in wrong range).
2) You have rounded the last digit incorrectly when giving the answer to 4 significant figures.

If working in SI units, what do you think the unit would be? (If I were handing this in, I would make a comment about the appropriate SI units below my answer.)
 
  • #7
Eitan Levy
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Agreed. Well done! A couple of points:
1) You haven't correctly used standard form (mantissa in wrong range).
2) You have rounded the last digit incorrectly when giving the answer to 4 significant figures.

If working in SI units, what do you think the unit would be? (If I were handing this in, I would make a comment about the appropriate SI units below my answer.)
[itex](Angstram)^{-1}[/itex]
 
  • #8
Steve4Physics
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[itex](Angstram)^{-1}[/itex]
The ångström (Å) (note spelling and capitalisation) is a bit old-fashioned, but is still used in some areas. It is not an SI unit.

1 Å = 10⁻¹⁰ m
That means
1 Å⁻¹ =10¹⁰ m⁻¹

I would have given the answer as:
α = 8.76x10⁸ (m⁻¹ assuming we are working in SI units)

EDIT: It's very poor practice to give values without units. The question should have given the units for k and b.
 

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