Question

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?

Answer 1

`lambda = 1.1 xx 10^(-38) "m"`

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The de Broglie wavelength `lambda` in `"m"` is given by the de Broglie relation:

`lambda = h/(mv)`

where:

• `h = 6.626 xx 10^(-34) "J" cdot"s"` is Planck's constant.
• `m` is the mass of the mass-ive object in `"kg"`.
• `v` is its velocity in `"m/s"`.

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

`"2.2 lb"/"kg"" "" "" ""5280 ft"/"1 mi"" "" ""12 in"/"1 ft"`

`"2.54 cm"/"1 in"" "" ""100 cm"/"1 m"" "" ""60 min"/"1 hr"`

`"60 s"/"1 min"`

First, convert the mass:

`5400 cancel"lbs" xx "1 kg"/(2.2 cancel"lbs") = "2454.5 kg"`
`" "" "" "" "" "" "" "" "`(we'll round later)

Now convert the velocity:

`(57 cancel"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")`

`=` `"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"))`

`= ulcolor(blue)(1.1 xx 10^(-38) "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.