# A certain sports utility vehicle is traveling at a speed of "57 mph". If its mass is "5400 lbs", what is its de Broglie wavelength?

Sep 7, 2017

$\lambda = 1.1 \times {10}^{- 38} \text{m}$

The de Broglie wavelength $\lambda$ in $\text{m}$ is given by the de Broglie relation:

$\lambda = \frac{h}{m v}$

where:

• $h = 6.626 \times {10}^{- 34} \text{J" cdot"s}$ is Planck's constant.
• $m$ is the mass of the mass-ive object in $\text{kg}$.
• $v$ is its velocity in $\text{m/s}$.

The units here are wacky, so we'll need to convert the $\text{lb}$ into $\text{kg}$ and the $\text{mph}$ into $\text{m/s}$. We have the following conversion factors to consider:

$\text{2.2 lb"/"kg"" "" "" ""5280 ft"/"1 mi"" "" ""12 in"/"1 ft}$

$\text{2.54 cm"/"1 in"" "" ""100 cm"/"1 m"" "" ""60 min"/"1 hr}$

$\text{60 s"/"1 min}$

First, convert the mass:

$5400 \cancel{\text{lbs" xx "1 kg"/(2.2 cancel"lbs") = "2454.5 kg}}$
$\text{ "" "" "" "" "" "" "" }$(we'll round later)

Now convert the velocity:

$\left(57 \cancel{\text{mi")/cancel"hr" xx (5280 cancel"ft")/(cancel"1 mi") xx (12 cancel"in")/(cancel"1 ft") xx (2.54 cancel"cm")/(cancel"1 in") xx "1 m"/(100 cancel"cm") xx (cancel"1 hr")/(60 cancel"min") xx (cancel"1 min")/("60 s}}\right)$

$=$ $\text{25.481 m/s}$
(we'll round later)

Lastly, just solve the de Broglie relation for the wavelength.

color(blue)(lambda) = (6.626 xx 10^(-34) cancel"kg" cdot"m"^cancel(2)"/"cancel"s")/(2454.5 cancel"kg" cdot 25.481 cancel("m/s"))

$= \underline{\textcolor{b l u e}{1.1 \times {10}^{- 38} \text{m}}}$

And this number is justifiably puny.

A macroscopic particle like a sports utility vehicle has practically no wave characteristics to speak of. Only really light particles moving at very fast speeds are quantum mechanical.