How Is Photon Energy Calculated When an Electron Moves to a Lower Energy State?

[mateomy]

An electron is trapped in an infinite one-dimensional well of width 0.251nm. Initially the electron occupies the n=4 state. Suppose the electron jumps to the ground state with the accompanying emission of a photon. What is the energy of the photon?

(Time independent)

What I did was realize that $\psi(x)$ must equal zero at the walls, so I chose $\sin(kL)$ and set it to $n\pi$, then solved for k.

Putting this value of k into the energy for a free particle (Time-ind Schrodinger), I eventually come to the expression:
$$\frac{h^2 n^2}{8mL^2}$$

To find the corresponding energy of the emitted photon I plugged in the appropriate values of n and solved for the difference.

Does that seem right?

Answer:
$$E=\frac{h^2 15}{8m L^2}$$

Obviously plugging in appropriate values of m and L afterward.

 

[Simon Bridge]

To find the corresponding energy of the emitted photon I plugged in the appropriate values of n and solved for the difference.

Does that seem right?

That is certainly the method.
If this is long answer you want to write it explicitly in the math though.

 

[mateomy]

Yeah, I have it officially written out showing steps and whatnot. Just felt lazy with the LaTex so I shortened it. Thanks.