Electron in 0.1 nm well: ground energy

For an electron in an infinite well the allowed energies are $E_n = \dfrac{n^2h^2}{8mL^2}$.

Given: $L = 0.1\ \text{nm} = 1\times10^{-10}$ m, $m = 9.1\times10^{-31}$ kg, $h = 6.626\times10^{-34}$ J s, $\hbar = h/2\pi = 1.055\times10^{-34}$ J s.

Ground state energy ($n = 1$)

$$\begin{aligned} E_1 &= \frac{h^2}{8mL^2} = \frac{(6.626\times10^{-34})^2}{8\times9.1\times10^{-31}\times(10^{-10})^2} \\ &= \frac{4.390\times10^{-67}}{7.28\times10^{-50}} = 6.03\times10^{-18}\ \text{J} \\ &= \frac{6.03\times10^{-18}}{1.6\times10^{-19}} = 37.7\ \text{eV} \end{aligned}$$

Radian frequency

$$\begin{aligned} \omega_1 &= \frac{E_1}{\hbar} = \frac{6.03\times10^{-18}}{1.055\times10^{-34}} \\ &= 5.72\times10^{16}\ \text{rad/s} \end{aligned}$$

Putting the electron in the third level

$$\begin{aligned} E_3 &= 3^2E_1 = 9\times37.7 = 339.2\ \text{eV} \\ \Delta E &= E_3 - E_1 = 8E_1 = 301.5\ \text{eV} = 4.82\times10^{-17}\ \text{J} \end{aligned}$$

The electron must absorb exactly this energy. It can be given by a photon of wavelength

$$\lambda = \frac{hc}{\Delta E} = \frac{6.626\times10^{-34}\times3\times10^{8}}{4.82\times10^{-17}} = 4.12\times10^{-9}\ \text{m} = 4.12\ \text{nm}$$

(a soft X-ray photon), or by a collision with another particle that transfers 301.5 eV.

Answer: $E_1 = 6.03\times10^{-18}$ J $= 37.7$ eV; $\omega = 5.72\times10^{16}$ rad/s; to reach $n = 3$ the electron must absorb $301.5$ eV, e.g. a photon of $\lambda \approx 4.12$ nm.