---
title: "Quantum mechanics and golf ball wavelength"
subject: "Electrical Engineering Material"
chapter: 1
key: "semester_3/electrical_engineering_material:golf-ball-de-broglie-wavelength"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/golf-ball-de-broglie-wavelength
site: BCT Engineering
---

# Quantum mechanics and golf ball wavelength

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2081 Baisakh, 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/golf-ball-de-broglie-wavelength
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

What do you mean by quantum mechanics? Evaluate the wavelength of 50 gram golf ball travelling at a velocity of 20 ms⁻¹.

## Short answer

Quantum mechanics describes wave–particle duality; de Broglie wavelength λ = h/(mv). For m = 0.05 kg, v = 20 m/s, λ = 6.626×10⁻³⁴ / (0.05×20) = 6.63×10⁻³⁴ m, far too small to detect.

## Answer

**Quantum mechanics** is the theory that describes the behaviour of very small particles (electrons, atoms, photons). In it, a particle is described by a wave function $\psi$, its energy is quantized when it is bound, and only probabilities of position and momentum can be predicted. A key idea is **wave–particle duality**: every moving particle has a de Broglie wavelength

$$
\lambda = \frac{h}{p} = \frac{h}{mv}
$$

#### Wavelength of the golf ball
**Given:** $m = 50\ \text{g} = 0.05$ kg, $v = 20$ m/s, $h = 6.626\times10^{-34}$ J s.

$$
\begin{aligned}
p &= mv = 0.05\times20 = 1\ \text{kg m s}^{-1} \\
\lambda &= \frac{h}{mv} = \frac{6.626\times10^{-34}}{1} = 6.626\times10^{-34}\ \text{m}
\end{aligned}
$$

**Answer:** $\lambda = 6.63\times10^{-34}$ m.

#### Meaning
This wavelength is more than $10^{18}$ times smaller than the size of a proton ($\sim10^{-15}$ m). No slit or crystal is small enough to show diffraction of the golf ball, so its wave nature can never be observed and classical mechanics describes it perfectly. For comparison, an electron accelerated through 100 V has $\lambda \approx 0.123$ nm, similar to the spacing between atoms, so electrons show clear diffraction and must be treated by quantum mechanics.

## Check yourself

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

1. Which relation gives the wavelength of a moving particle?
   A. lambda = mv/h
   B. lambda = h/(mv)
   C. lambda = hv/m
   D. lambda = h m v
   Answer: B. de Broglie wavelength is lambda = h/p = h/(mv). (Key point: de Broglie relation)

2. What is the wavelength of the 50 g golf ball at 20 m/s?
   A. 6.6e-34 m
   B. 6.6e-31 m
   C. 6.6e-36 m
   D. 6.6e-15 m
   Answer: A. p = 0.05 x 20 = 1 kg m/s, so lambda = 6.626e-34 / 1 = 6.63e-34 m. (Key point: Calculation)

3. What does this tiny wavelength tell us?
   A. The golf ball is a quantum particle
   B. Its wave nature can never be observed, so classical mechanics suffices
   C. It should show diffraction in crystals
   D. It equals the atomic spacing
   Answer: B. The wavelength is far smaller than any slit or atom, so no diffraction is observable. (Key point: Meaning)

## Related in Electrical Engineering Material

- [Classical versus quantum mechanics](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aclassical-vs-quantum-mechanics) — Electrical Engineering Material, chapter 1, asked 2×

## Related in other subjects

- [Ball tossed upward from window; dependent motion](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aball-tossed-vertical-window-dependent-motion) — Applied Mechanics, chapter 8
- [Field channel discharge for gram and wheat](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Afield-channel-gram-wheat-2000ha) — Irrigation and Drainage Engineering, chapter 2
- [Ball strikes suspended rod, constrained motion](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Aball-strikes-suspended-rod-50n) — Applied Mechanics (Dynamics), chapter 6
- [Settling tank 20 m x 2.5 m; velocity for 0.02 mm](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-b66) — Water Supply Engineering, chapter 6, asked 2×