Electron in 0.01 nm well: third level

For an electron in a one-dimensional infinite potential well of width $L$, the allowed energies and normalised wave functions are

$$E_n = \frac{n^2 h^2}{8 m L^2}, \qquad \psi_n(x) = \sqrt{\frac{2}{L}}\,\sin\left(\frac{n\pi x}{L}\right),\quad 0 \le x \le L$$

and $\psi_n = 0$ outside the well.

Given: $L = 0.01\ \text{nm} = 1 \times 10^{-11}\ \text{m}$, $n = 3$, $h = 6.626 \times 10^{-34}\ \text{J s}$, $m = 9.109 \times 10^{-31}\ \text{kg}$.

Energy at third level

$$\begin{aligned} E_3 &= \frac{3^2 h^2}{8 m L^2} = \frac{9 \times (6.626 \times 10^{-34})^2}{8 \times 9.109 \times 10^{-31} \times (10^{-11})^2} \\ &= 5.42 \times 10^{-15}\ \text{J} \\ &= \frac{5.42 \times 10^{-15}}{1.602 \times 10^{-19}} = 3.38 \times 10^{4}\ \text{eV} \approx 33.8\ \text{keV} \end{aligned}$$

Wave function at third level

$$\sqrt{\frac{2}{L}} = \sqrt{\frac{2}{10^{-11}}} = 4.47 \times 10^{5}\ \text{m}^{-1/2}, \qquad \frac{3\pi}{L} = 9.42 \times 10^{11}\ \text{m}^{-1}$$ $$\psi_3(x) = 4.47 \times 10^{5}\, \sin\left(9.42 \times 10^{11}\, x\right)\ \text{m}^{-1/2}, \quad 0 \le x \le 10^{-11}\ \text{m}$$

$\psi_3$ has two nodes inside the well (at $x = L/3$ and $2L/3$) and three half-wavelengths fit in the well.

Answer: $E_3 = 5.42 \times 10^{-15}\ \text{J} \approx 33.8\ \text{keV}$; $\psi_3(x) = 4.47 \times 10^{5} \sin(9.42 \times 10^{11} x)$ inside the well, zero outside.