---
title: "Electron in 0.01 nm well: third level"
subject: "Electrical Engineering Material"
chapter: 1
key: "semester_3/electrical_engineering_material:well-third-level-wave-function"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/well-third-level-wave-function
site: BCT Engineering
---

# Electron in 0.01 nm well: third level

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2071 Shrawan, 4 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-third-level-wave-function
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

An electron is confined in an infinite potential well. The length of confinement is 0.01 nm. Find the energy and wave function of electron at third energy level.

## Short answer

Use E_n = n²h²/(8mL²) with L=0.01 nm and n=3. E_3 = 5.42×10^{-15} J ≈ 33.8 keV. The normalized wavefunction is ψ_3(x) = √(2/L) sin(3πx/L) = 4.47×10^5 sin(9.42×10^{11} x) m^{-1/2} inside the well, zero outside.

## Answer

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.

## Check yourself

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

1. Which formula gives the third-level energy in an infinite well?
   A. E3 = 3 h^2 / 8mL^2
   B. E3 = 9 h^2 / 8mL^2
   C. E3 = 9 h^2 / 2mL
   D. E3 = h^2 / 72 mL^2
   Answer: B. E_n = n^2 h^2 / 8mL^2 with n = 3. (Key point: Energy formula)

2. For L = 0.01 nm, E3 is about:
   A. 33.8 eV
   B. 3.38 keV
   C. 33.8 keV
   D. 338 keV
   Answer: C. E3 = 5.42e-15 J = 3.38e4 eV. (Key point: Energy result)

3. Which is the normalised wave function for n = 3?
   A. sqrt(2/L) cos(3 pi x/L)
   B. sqrt(2/L) sin(3 pi x/L)
   C. sqrt(L/2) sin(3 pi x/L)
   D. sqrt(2/L) sin(pi x/3L)
   Answer: B. psi_n = sqrt(2/L) sin(n pi x/L), zero at the walls. (Key point: Wave function)

4. How many nodes does psi_3 have inside the well?
   A. 1
   B. 2
   C. 3
   D. 4
   Answer: B. n - 1 = 2 nodes, at L/3 and 2L/3. (Key point: Nodes)

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