---
title: "Electron in 0.1 nm well: ground energy"
subject: "Electrical Engineering Material"
chapter: 1
key: "semester_3/electrical_engineering_material:well-numerical-ground-energy"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/well-numerical-ground-energy
site: BCT Engineering
---

# Electron in 0.1 nm well: ground energy

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2081 Bhadra, 4 marks; 2075 Chaitra, 4 marks; 2072 Chaitra, 4 marks (asked 3 times)
- 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-ground-energy
- 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 0.1nm. Calculate the ground energy of the electron and radian frequency. How this electron can be put to the third energy level?

## Short answer

Using E_n = n²h²/(8mL²) for L=0.1 nm, ground energy E₁ = 6.03×10⁻¹⁸ J = 37.7 eV, radian frequency ω = E₁/ħ = 5.72×10¹⁶ rad/s. To reach n=3, the electron must absorb ΔE = 301.5 eV, e.g., via a photon of wavelength ≈ 4.12 nm.

## Answer

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.

## Check yourself

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

1. Which formula gives the ground energy of an electron in an infinite well of width L?
   A. E1 = h^2 / (2 m L^2)
   B. E1 = h^2 / (8 m L^2)
   C. E1 = h / (8 m L^2)
   D. E1 = 8 h^2 / (m L^2)
   Answer: B. E_n = n^2 h^2 / (8 m L^2) with n = 1. (Key point: Formula used)

2. For L = 0.1 nm, what is the ground state energy?
   A. 3.77 eV
   B. 377 eV
   C. 37.7 eV
   D. 0.377 eV
   Answer: C. E1 = 6.03e-18 J = 37.7 eV. (Key point: Ground energy value)

3. How much energy must the electron absorb to go from n = 1 to n = 3?
   A. 8 E1 = 301.5 eV
   B. 9 E1 = 339.2 eV
   C. 3 E1 = 113 eV
   D. 2 E1 = 75.4 eV
   Answer: A. Delta E = E3 - E1 = 9E1 - E1 = 8E1 = 301.5 eV, absorbed from a photon or a collision. (Key point: Reaching n = 3)

## Close variants set in other exams

- [Electron in 8.5 nm well: fourth level](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Awell-numerical-8nm-fourth-level) — Electrical Engineering Material, chapter 1, asked 4×

## Related in Electrical Engineering Material

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- [Infinite potential well: quantized energy and wave function](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Ainfinite-well-energy-quantized) — Electrical Engineering Material, chapter 1, asked 4×
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