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Hello,
It's only been recently that I have acquired the math skills to deal with the time independent version of Schrodinger's Equation which is:
[tex] \frac{-\hbar^2}{2m} \frac{d^2}{dx^2}\Psi(x) + U(x)\Psi(x) = E\Psi(x)[/tex]
I tried to derive a wavefunction that deals with a particle in a confined box with infinite walls as shown http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/pbox.html#c1"
I was able to get something similar to:
[tex] \Psi(x) = A*sin(kx)[/tex]
What somewhat baffles me is how they define lambda( the wavelength within k as shown http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/schr.html#c2"). Perhaps I am missing the obvious but why would the biggest wavelength be
[tex] \lambda = 2L [/tex]
Why not have
[tex] \lambda = 4L [/tex]
Or
[tex] \lambda = 7L [/tex]
What makes [tex] \lambda = 2L [/tex] the biggest wavelength value. This would clear up a ton.
The second and last question I have(so far) is what particular value of n does a particle need to take? What does the value of n depend upon?
Thanks!
It's only been recently that I have acquired the math skills to deal with the time independent version of Schrodinger's Equation which is:
[tex] \frac{-\hbar^2}{2m} \frac{d^2}{dx^2}\Psi(x) + U(x)\Psi(x) = E\Psi(x)[/tex]
I tried to derive a wavefunction that deals with a particle in a confined box with infinite walls as shown http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/pbox.html#c1"
I was able to get something similar to:
[tex] \Psi(x) = A*sin(kx)[/tex]
What somewhat baffles me is how they define lambda( the wavelength within k as shown http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/schr.html#c2"). Perhaps I am missing the obvious but why would the biggest wavelength be
[tex] \lambda = 2L [/tex]
Why not have
[tex] \lambda = 4L [/tex]
Or
[tex] \lambda = 7L [/tex]
What makes [tex] \lambda = 2L [/tex] the biggest wavelength value. This would clear up a ton.
The second and last question I have(so far) is what particular value of n does a particle need to take? What does the value of n depend upon?
Thanks!
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