Quantum mechanics - several constant potentials

AI Thread Summary
The discussion focuses on solving equations related to quantum mechanics involving wave functions and their continuity. Participants emphasize the importance of using the first two equations for simplification and express concerns about the complexity of the calculations. A specific value of 0.08763*10^(10) is questioned, with discussions about the absence of units in the original problem. Clarifications are made regarding the use of standard form and the importance of including appropriate SI units in calculations. Overall, the conversation highlights the need for precision in both calculations and unit representation in quantum mechanics problems.
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
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What I tried to do was using the fact that the wave function should be continuous.

Asin(kb)=Be^{-\alpha b}

The derivative also should be continuous:

kAcos(kb)=-\alpha Be^{-\alpha b}

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

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

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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Eitan Levy said:
Asin(kb)=Be^{-\alpha b}

kAcos(kb)=-\alpha Be^{-\alpha b}

\int_0^b A^2sin^2(kx) dx + \int_b^{\infty} B^2e^{-2\alpha x} dx =1
Hints: It's much easier than you think. You only need the first 2 equations.
 
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Steve4Physics said:
Hints: It's much easier than you think. You only need the first 2 equations.
Is the answer 0.08763*10^(10)?
 
Eitan Levy said:
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?
 
Steve4Physics said:
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:

\frac{1}{k}tan(kb)=-\frac{1}{\alpha}

And I plugged in the numbers.

No units were provided.
 
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Eitan Levy said:
I divided the first equation by the second:

\frac{1}{k}tan(kb)=-\frac{1}{\alpha}

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.)
 
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Steve4Physics said:
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.)
(Angstram)^{-1}
 
Eitan Levy said:
(Angstram)^{-1}
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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