---
title: "Tunneling probability of a ball over barrier"
subject: "Electrical Engineering Material"
chapter: 1
key: "semester_3/electrical_engineering_material:tunneling-ball-barrier"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/tunneling-ball-barrier
site: BCT Engineering
---

# Tunneling probability of a ball over barrier

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2080 Bhadra, 4 marks
- Video: [Electron tunneling through a barrier](https://www.youtube.com/watch?v=8-pBXRQIq6M) — Winfernity
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/tunneling-ball-barrier
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

Estimate the probability of transmission that a ball weighing 0.5 g released at a height of 5 m at rest will reach a barrier height of 8 m with barrier width of 1.5 m.

## Short answer

Ball mass 0.5 g: E = mgh₁ = 0.0245 J, V₀ = 0.0392 J. α = √(2m(V₀−E))/ħ = 3.64×10³¹ m⁻¹, T ≈ 16(E/V₀)(1−E/V₀)e^(−2αa) ≈ 3.75e^(−1.09×10³²) ≈ 0. Practically zero transmission.

## Answer

A particle of energy $E$ meeting a barrier of height $V_0 > E$ and width $a$ can tunnel through with probability

$$
T \approx T_0\,e^{-2\alpha a}, \qquad \alpha = \frac{\sqrt{2m(V_0 - E)}}{\hbar}, \qquad T_0 = 16\frac{E}{V_0}\left(1 - \frac{E}{V_0}\right)
$$

**Given:** $m = 0.5\ \text{g} = 5\times10^{-4}$ kg, release height 5 m, barrier height 8 m, width $a = 1.5$ m, $g = 9.81$ m/s², $\hbar = 1.055\times10^{-34}$ J s.

#### Energies
The ball released from rest at 5 m has total energy (its kinetic energy at ground level):

$$
E = mgh_1 = 5\times10^{-4}\times9.81\times5 = 0.02453\ \text{J}
$$

To climb an 8 m barrier it would need

$$
V_0 = mgh_2 = 5\times10^{-4}\times9.81\times8 = 0.03924\ \text{J}
$$

$$
V_0 - E = 0.01472\ \text{J}
$$

Classically, $E < V_0$, so the ball can never cross.

#### Decay constant

$$
\begin{aligned}
\alpha &= \frac{\sqrt{2\times5\times10^{-4}\times0.01472}}{1.055\times10^{-34}} = \frac{3.836\times10^{-3}}{1.055\times10^{-34}} \\
&= 3.64\times10^{31}\ \text{m}^{-1}
\end{aligned}
$$

$$
2\alpha a = 2\times3.64\times10^{31}\times1.5 = 1.09\times10^{32}
$$

#### Transmission probability

$$
T_0 = 16\times\frac{0.02453}{0.03924}\left(1 - \frac{0.02453}{0.03924}\right) = 16\times0.625\times0.375 = 3.75
$$

$$
\begin{aligned}
T &\approx 3.75\,e^{-1.09\times10^{32}} \\
&\approx 10^{-4.74\times10^{31}} \approx 0
\end{aligned}
$$

**Answer:** $T \approx e^{-1.09\times10^{32}} \approx 10^{-4.7\times10^{31}}$, which is zero for all practical purposes.

The ball will never be seen tunneling through the barrier, because its mass is huge compared with an electron and the barrier is very wide. Tunneling matters only for very light particles and barriers of nanometre width.

## Check yourself

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

1. How are E and V0 obtained for the ball?
   A. E = mg h1 (5 m) and V0 = mg h2 (8 m)
   B. E = mg h2 and V0 = mg h1
   C. Both equal mgh with h = 1.5 m
   D. E = 0 and V0 = mg x 8
   Answer: A. Total energy equals the potential energy at 5 m; the barrier corresponds to the 8 m height, so E < V0. (Key point: Energies)

2. Which expression gives alpha?
   A. sqrt(2mE)/hbar
   B. sqrt(2m(V0 - E))/hbar
   C. 2m(V0 - E)/hbar^2
   D. hbar/sqrt(2m(V0 + E))
   Answer: B. alpha = sqrt(2m(V0 - E))/hbar, here about 3.64e31 per metre. (Key point: Decay constant)

3. What is the conclusion for the ball's transmission probability?
   A. About 50%
   B. About 1%
   C. T = 3.75, greater than 1
   D. T is e^(-1.09e32), essentially zero
   Answer: D. The huge exponent 2 alpha a makes T vanishingly small because the ball is massive and the barrier wide. (Key point: Result)

## Related in Electrical Engineering Material

- [Derivation of tunneling probability through barrier](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Atunneling-probability-derivation) — Electrical Engineering Material, chapter 1

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