Electron in 8.5 nm well: fourth level

Given: $L = 8.5\ \text{nm} = 8.5 \times 10^{-9}\ \text{m}$, $m = 9.109 \times 10^{-31}\ \text{kg}$, $h = 6.626 \times 10^{-34}\ \text{J s}$, $\hbar = h/2\pi = 1.0546 \times 10^{-34}\ \text{J s}$.

Ground state energy (n = 1)

$$\begin{aligned} E_1 &= \frac{h^2}{8 m L^2} = \frac{(6.626 \times 10^{-34})^2}{8 \times 9.109 \times 10^{-31} \times (8.5 \times 10^{-9})^2} \\ &= 8.34 \times 10^{-22}\ \text{J} = 5.21 \times 10^{-3}\ \text{eV} \end{aligned}$$

Radian frequency

From $E = \hbar\omega$:

$$\omega_1 = \frac{E_1}{\hbar} = \frac{8.34 \times 10^{-22}}{1.0546 \times 10^{-34}} = 7.91 \times 10^{12}\ \text{rad/s}$$

Raising the electron to the fourth level

$E_4 = 4^2 E_1 = 16E_1$, so the electron must absorb energy

$$\Delta E = E_4 - E_1 = 15E_1 = 15 \times 5.21 \times 10^{-3} = 0.0781\ \text{eV}\ (1.25 \times 10^{-20}\ \text{J})$$

This can be given by a photon of exactly this energy:

$$\lambda = \frac{hc}{\Delta E} = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{1.25 \times 10^{-20}} = 1.59 \times 10^{-5}\ \text{m} = 15.9\ \mu\text{m}$$

i.e. infrared radiation of frequency $f = \Delta E/h = 1.89 \times 10^{13}\ \text{Hz}$. (The energy could also come from a collision, but it must equal exactly $15E_1$; energy in between is not accepted.)

Answer: $E_1 = 8.34 \times 10^{-22}\ \text{J} = 5.21\ \text{meV}$; $\omega = 7.91 \times 10^{12}\ \text{rad/s}$; to reach $n = 4$ it must absorb $0.0781\ \text{eV}$ (a photon of $\lambda \approx 15.9\ \mu\text{m}$).