---
title: "Electron in 8.5 nm well: fourth level"
subject: "Electrical Engineering Material"
chapter: 1
key: "semester_3/electrical_engineering_material:well-numerical-8nm-fourth-level"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/well-numerical-8nm-fourth-level
site: BCT Engineering
---

# Electron in 8.5 nm well: fourth level

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2081 Chaitra (new course), 3 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/well-numerical-8nm-fourth-level
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

An electron is confined to an infinite potential well of size 8.5 nm. Calculate the ground state energy of the electron and radian frequency. How this electron can be put to the fourth energy level?

## Short answer

For an infinite well of width L=8.5 nm, ground state energy E1 = h²/(8mL²) = 8.34×10⁻²² J (5.21 meV) and radian frequency ω = E1/ħ = 7.91×10¹² rad/s. To reach the fourth level (n=4), the electron must absorb ΔE = 15E1 = 0.0781 eV, e.g., a photon of λ ≈ 15.9 μm.

## Answer

**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 $\lambda \approx 15.9\ \mu\text{m}$).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is E1 for the 8.5 nm well?
   A. 8.34e-22 J (5.21 meV)
   B. 5.21 eV
   C. 8.34e-19 J
   D. 1.25e-20 J
   Answer: A. E1 = h^2/8mL^2 = 8.34e-22 J = 5.21e-3 eV. (Key point: Ground-state energy)

2. How is the radian frequency found?
   A. omega = E1 x hbar
   B. omega = E1/hbar
   C. omega = E1/h
   D. omega = hbar/E1
   Answer: B. E = hbar omega gives 7.91e12 rad/s. (Key point: Radian frequency)

3. What energy must be absorbed to go from n = 1 to n = 4?
   A. 16 E1
   B. 15 E1
   C. 4 E1
   D. 3 E1
   Answer: B. E4 - E1 = 16E1 - E1 = 15E1 = 0.0781 eV, supplied exactly, e.g. by a 15.9 micrometre photon. (Key point: Energy to reach n = 4)

## Close variants set in other exams

- [Electron in 0.1 nm well: ground energy](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Awell-numerical-ground-energy) — Electrical Engineering Material, chapter 1, asked 4×

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