# Compare the de Broglie wavelengths for a proton, golf ball, and alpha particle, if these have speeds of #5.81 xx 10^6 "m/s"#, #"31.3 m/s"#, and #1.52 xx 10^7 "m/s"#, respectively? Which one is a classical particle?

##### 1 Answer

The wavelengths for the proton (

#lambda_p > lambda_(alpha)# #">>"# #lambda_(gb)#

This physically means that among the objects in this list, the proton is most accurately described by **quantum mechanics**, followed by the alpha particle. They have significant wave characteristics.

The golf ball just fails catastrophically; it is clearly visible to the naked eye, and so it is described by **classical mechanics**, not quantum mechanics. It has practically zero wave characteristics.

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 the object'smassin#"kg"# and#v# is itsvelocityin#"m/s"# .

To compare

- The rest mass
#m_p# of a proton is#ul(1.673 xx 10^(-27) "kg")# . - The maximum mass of a regulation golf ball,
#m_(gb)# , is#"45.93 g"# , or#ul"0.04593 kg"# . - An
#alpha# particle, or a helium nucleus, has a mass#m_(alpha)# of#ul(6.646 xx 10^(-27) "kg")# , based on the mass of#"4.0026 amu"# for the helium element, converted to#"kg"# using the factor#1.6605 xx 10^(-24) "g/amu"# .

And so, we define the masses as:

#m_p = 1.673 xx 10^(-27) "kg"# #m_(gb) = "0.04593 kg"# #m_(alpha) = 6.646 xx 10^(-27) "kg"#

and the velocities were given as:

#v_p = 5.81 xx 10^6 "m/s"# #v_(gb) = "31.3 m/s"# #v_(alpha) = 1.52 xx 10^7 "m/s"#

The de Broglie wavelengths would then follow.

Note that the de Broglie relation assumes the particle is also a wave. Everything *has* a wave characteristic, BUT the idea is, is it *significant*? Let's find out...

**For the proton:**

#lambda_p = h/(m_p v_p) = (6.626 xx 10^(-34) "kg"cdot"m"^2"/s")/(1.673 xx 10^(-27) "kg" cdot 5.81 xx 10^6 "m/s")#

#= 6.82 xx 10^(-14) "m"# , or about#"68.2 fm"# (femtometers).

As the root-mean-square charge radius of a proton is about

**For the golf ball:**

#lambda_(gb) = h/(m_(gb) v_(gb)) = (6.626 xx 10^(-34) "kg"cdot"m"^2"/s")/("0.04593 kg" cdot "31.3 m/s")#

#= 4.61 xx 10^(-34) "m"# , or about#4.61 xx 10^(-19) "fm"# (VERY TINY!).

This number is ** not supposed** to make sense, because golf balls are described as macroparticles, i.e. they are the size of everyday objects. They don't exhibit the

**wave-like qualities**we see in quantum-sized objects like electrons and protons.

*So, the wavelength of a golf ball SHOULD be ridiculously small like this.*

**For the helium nucleus:**

#lambda_(alpha) = h/(m_(alpha) v_(alpha)) = (6.626 xx 10^(-34) "kg"cdot"m"^2"/s")/(6.646 xx 10^(-27) "kg" cdot 1.52 xx 10^7 "m/s")#

#= 6.56 xx 10^(-15) "m"# , or about#"6.56 fm"# (femtometers).

Comparing with the proton above, the wavelength for the proton is about

**In the end, we conclude that the lighter and slower the particle, the more of a wave characteristic it exhibits.**