One dimensional Schroedinger equation solution

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The discussion revolves around demonstrating that the wave function ψ(x) = B*sin(kx-∅) is a solution to the one-dimensional Schrödinger equation before a potential barrier, where U(x) = 0. The user attempts to derive the energy E and the wave number k by substituting the wave function into the equation. They find a relationship for k that appears to be a multiple of 2π/λ instead of the expected form. The user seeks clarification on their calculations and expresses gratitude for any assistance. The thread highlights the importance of correctly applying the Schrödinger equation in quantum mechanics.
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Homework Statement


before a potential barrier show;
ψ(x) = B*sin(kx-∅)
B, k and ∅ are all possitive

Is a solution of the one dimensional Schroedinger equation

Homework Equations


ψ(x) = B*sin(kx-∅)
Eψ(x) = -(ħ2/2m)(d2ψ(x)/dx2)+U(x)ψ(x)

The Attempt at a Solution


If it is before the barrier U(x) = 0 so;
Eψ(x) = -(ħ2/2m)(d2ψ(x)/dx2)

ψ(x) = B*sin(kx-∅)
dψ(x)/dx= Bk*cos(kx-∅)
(d2ψ(x)/dx2) = -Bk2*sin(kx-∅)

By substitution;
E = ħ22k2/2m
λ = h/p (de broglie wavelength)
p = mv
h = λmv
so k = 2π√(2Em) / λmv
??
apparently I should get;
k=2π/λ
What I have is a multiple of 2π/λ
What have I done wrong / what else should I do
Thanks in advance for and help you offer
 
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E = 0.5 * m * v^2 I got it :)
 
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