---
title: "Parabolic E-k and quantized linear metal"
subject: "Electrical Engineering Material"
chapter: 1
key: "semester_3/electrical_engineering_material:ek-parabolic-linear-metal"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/ek-parabolic-linear-metal
site: BCT Engineering
---

# Parabolic E-k and quantized linear metal

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2070 Chaitra, 4+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/ek-parabolic-linear-metal
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

From free electron theory of metal, show that E-K diagram is parabolic. Also show the energy of electron in a linear metal is quantized.

## Short answer

From free electron theory with V=0, Schrödinger’s equation yields E = ħ²k²/2m, a parabolic E–k curve. In a linear metal of length L, boundary conditions ψ(0)=ψ(L)=0 enforce k = nπ/L, giving quantized energies E_n = n²h²/8mL².

## Answer

#### E–k diagram is parabolic
In the **free electron theory**, the valence electrons of a metal move freely inside the metal; the potential inside is taken as constant ($V = 0$). All the energy is kinetic:

$$
E = \frac{1}{2}mv^2 = \frac{p^2}{2m}
$$

By de Broglie, $p = h/\lambda = \hbar k$ where $k = 2\pi/\lambda$ is the wave number:

$$
E = \frac{\hbar^2k^2}{2m}
$$

The same follows from the Schrödinger equation with $V = 0$: $\psi = Ae^{jkx}$ gives $-\frac{\hbar^2}{2m}(-k^2)\psi = E\psi$.

Since $\hbar^2/2m$ is constant, $E \propto k^2$: the E–k curve is a **parabola**, symmetric about $k = 0$.

```text
   E
   ^
   |  \                 /
   |   \               /     E = (hbar k)^2 / 2m
   |    \             /
   |     `.         .'
   |       `-.___.-'
   +------------|-------------> k
                0
```

The slope gives velocity $v = \frac{1}{\hbar}\frac{dE}{dk} = \frac{\hbar k}{m}$, and the curvature gives $m^* = m$.

#### Energy in a linear metal is quantized
Consider a one-dimensional (linear) metal of length $L$, e.g. a thin wire. Electrons move freely inside ($V = 0$) but cannot leave it, because of the large surface barrier (work function). So it is a potential well with $V = \infty$ outside.

Inside, the Schrödinger equation is

$$
\frac{d^2\psi}{dx^2} + k^2\psi = 0, \qquad k^2 = \frac{2mE}{\hbar^2}
$$

with solution $\psi = A\sin kx + B\cos kx$. Boundary conditions:
- $\psi(0) = 0 \Rightarrow B = 0$
- $\psi(L) = A\sin kL = 0 \Rightarrow kL = n\pi$, so $k_n = \dfrac{n\pi}{L}$, $n = 1, 2, 3, \ldots$

Therefore

$$
E_n = \frac{\hbar^2k_n^2}{2m} = \frac{n^2h^2}{8mL^2}
$$

Only certain values of $k$ (spaced $\pi/L$ apart) and energy are allowed: **the energy is quantized**. The parabola above is really a set of closely spaced points.

#### Spacing of levels
For $L = 1$ cm, $E_1 = \dfrac{(6.626\times10^{-34})^2}{8\times9.1\times10^{-31}\times(10^{-2})^2} = 6.03\times10^{-34}$ J $= 3.8\times10^{-15}$ eV. The levels are so close that in a macroscopic metal the energy appears continuous, but strictly it is quantized; quantization becomes important only when $L$ is in the nanometre range (quantum wires).

## Check yourself

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

1. Why does the free electron model have all of the electron's energy as kinetic energy?
   A. The electron is at rest
   B. The potential inside the metal is taken as constant (V = 0)
   C. The electron has no mass
   D. The potential is infinite inside
   Answer: B. Inside the metal V is taken as zero, so E = p^2/2m. (Key point: Free electron energy)

2. Using p = hbar k, which relation shows the E-k curve is a parabola?
   A. E = hbar k / 2m
   B. E = hbar^2 k^2 / 2m, so E is proportional to k^2
   C. E = 2m / (hbar k)^2
   D. E = hbar k^3 / m
   Answer: B. E = p^2/2m with p = hbar k gives E proportional to k^2, a parabola symmetric about k = 0. (Key point: Parabolic E-k)

3. In the 1-D metal of length L, what condition leads to k_n = n pi / L?
   A. psi(L) = 0 gives sin kL = 0 after psi(0) = 0 removed the cosine term
   B. psi is maximum at both ends
   C. The electron has zero energy at the ends
   D. k must be an integer
   Answer: A. psi(0) = 0 gives B = 0, and psi(L) = A sin kL = 0 requires kL = n pi. (Key point: Boundary conditions in linear metal)

4. What is the energy of an electron in the linear metal?
   A. E_n = n h^2 / 8mL
   B. E_n = n^2 h^2 / 8mL^2
   C. E_n = h^2 / 8mL^2 for all n
   D. E_n = n^2 h / 8mL^2
   Answer: B. E_n = hbar^2 k_n^2 / 2m = n^2 h^2 / 8mL^2, only discrete values. (Key point: Quantised energy)

5. Why does energy look continuous in a macroscopic metal?
   A. It is not quantised at all
   B. Levels are spaced extremely closely when L is large
   C. Electrons have no wave nature
   D. Temperature removes quantisation
   Answer: B. For L = 1 cm the spacing is tiny; quantisation matters only for nanometre sizes. (Key point: Seeming continuity)

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