# 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?

##### 1 Answer

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

The de Broglie wavelength **de Broglie relation**:

#lambda = h/(mv)# where:

#h = 6.626 xx 10^(-34) "J" cdot"s"# isPlanck's constant.#m# is themassof the mass-ive object in#"kg"# .#v# is itsvelocityin#"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"#

#"2.54 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.*