Question #243d0

1 Answer
Truong-Son N.
Oct 1, 2017

Tiny, insignificant, and unimportant to quantum mechanics.

Any mass-ive object follows the de Broglie relation:

#lambda = h/(mv)#,

where:

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

And so, knowing the mass is #"0.200 kg"#, we ought to convert the speed...

#v = "5 m"/cancel"hr" xx cancel"1 hr"/(60 cancel"min") xx cancel"1 min"/"60 s"#

#=# #"0.00139 m/s"#

Note... what golf ball even is a whopping #"200 g"#? A regulation golf ball is at most #"45.93 g"#. This is a sad overweight golf ball that was never meant to be...

Its wavelength should be insignificant...

#color(blue)(lambda) = (6.626 xx 10^(-34) cancel"kg"cdot"m"^cancel(2)"/"cancel"s")/("0.200"cancel"kg" xx "0.00139 "cancel"m/s")#

#= color(blue)(4.60 xx 10^(-36) "m")#

which as expected, is not worth mentioning. Usual wavelengths for atomic-sized objects are over #10^(27)# times longer.

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