I'm not solving **** until they say why they're allowing the electron potential to be infinity after L instead of a rational number based on a function of distance.
Why would you not solve that? It makes it much more simple, since you can assume the electron can not be anywhere else than in side the well.
You do this first, and after that you start to assume that the particle can actually be outside the well, and start working from there.
This is really rational in quantum physics, you start by assuming simple stuff and work your way up.
This is just a simple infinite well problem in quatum mechanics.
The solution is
sqrt(2/L)sin((n*Pi*x)/L)
Where L is the width, n is the state/mode, and x is the position in the well.
This is really simple stuff. Really just a solution to the Schrödinger equation.
Deleted comment coz i made a mistake:
SO ill start over with out my mistake:
The question asks (basically) to prove, using the equation shown, that, for the set of value given (where X is greater than 0 but smaller than L), there is a difference of one node between each solution to the equation.
i could prove to you how, but ill aboid it coz it would take me a while, but the solution is, as said by PenguinsOfMars:
(Square root of (2/L)) * sin of (n * Pi * X)/L))
L stands for the length of the box in question.
n is the state of the electron, that is, simply put, how much energy it has.
x is the position of the electron within the box.
Hopefully i havent made any more mistakes and hope that i helped you.
Have a nice day with your new physics knowledge
P.S. Feel free to correct a mistake i made.
P.S.S. imho PenguinsOfMars made a very straight forward and nice explanation. Props to you sir/mam
I know how to solve the equation, but I don't remember any "node" terms involved.
Here is the solution:
When V = infinite the wave function has to be zero because there is 0 probability of it being found in that region. When V=0 then it doesn't matter what the wave funciton is because V(x) * W = 0. We set the boundary to be W(0)=W(L)=0 (like i said, W has to be zero when V is infinite).
The equation simplifies to:
-h^2 / 2m * Laplacian(W) = i*h d/dt (W) = E*W(x)
We then let k = sqrt ( 2mE/h^2)
now we rewrite the PDE as an ODE:
d^2 / dx^2 (W(x)) + k^2 W(x) = 0
and to determine A we plug that bad boy into the ODE and we should get
Wn = Sqrt(2/L) * Sin(n*pi*x/L) when x is in (0,L) and it's nothing (as stated in the first comment) otherwise.
Not sure what nodes are but I guess since x is between 0 and L we are guaranteed that we don't get an integer multiple of pi at the endpoints but we do get some in between and I guess with a larger n we would have a larger interval with zeros, which i think are nodes (a guess). In other words, Wn-1 will have less zeros /node(?) than W_n.
A semester is a half a year. Midterms and finals end at the end of the semesters. A quarter is a quarter of a year, obviously. You would have quarterlies at the end of every quarter.
Google the "electron in a box" model. The Schroedinger equation describes how the oscillations of an electron can occur at certain harmonic frequencies with wavelengths L, L/2, L/3, L/4, etc. It's just quantum mechanics.