Quantum mechanics and golf ball wavelength

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.