A certain sports utility vehicle is traveling at a speed of "57 mph"57 mph. If its mass is "5400 lbs"5400 lbs, what is its de Broglie wavelength?
1 Answer
lambda = 1.1 xx 10^(-38) "m"λ=1.1×10−38m
The de Broglie wavelength
lambda = h/(mv)λ=hmv where:
h = 6.626 xx 10^(-34) "J" cdot"s"h=6.626×10−34J⋅s is Planck's constant.mm is the mass of the mass-ive object in"kg"kg .vv is its velocity in"m/s"m/s .
The units here are wacky, so we'll need to convert the
"2.2 lb"/"kg"" "" "" ""5280 ft"/"1 mi"" "" ""12 in"/"1 ft"2.2 lbkg 5280 ft1 mi 12 in1 ft
"2.54 cm"/"1 in"" "" ""100 cm"/"1 m"" "" ""60 min"/"1 hr"2.54 cm1 in 100 cm1 m 60 min1 hr
"60 s"/"1 min"60 s1 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.