Answer to Question #152571 in Quantum Mechanics for sami

Question

Show that the de Broglie wavelength of a particle in a one dimensional box in the first
excited state is equal to the length of the box.

Answer 1

Energy of the particle in a one-dimensional box

$E=\frac{n^2\pi^2\hbar^2}{2mL^2}=\frac{n^2\pi^2h^2}{8\pi^2mL^2}=\frac{n^2h^2}{8mL^2}$

Assume that $E=\frac{p^2}{2m}$.

$\frac{p^2}{2m}=\frac{n^2h^2}{8mL^2}\to L^2=\frac{n^2h^2}{4p^2}$

For $n=2$

$L=\frac{h}{p}$

The de Broglie wavelength

$\lambda=\frac{h}{p}$

So, $\lambda=L$ . Answer