Question

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?

Answer 1

The wavelengths for the proton (`p`), alpha particle (`alpha`), and golf ball (`gb`) compare as:

`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 `lambda` in units of `"m"` for mass-ive objects 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 object's mass in `"kg"` and `v` is its velocity in `"m/s"`.

To compare `lambda` for each object, first calculate them. You would have to already know their masses.

• 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 `"0.87 fm"`, this wavelength makes physical sense.

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 `10` times the wavelength for this `alpha` particle at this velocity, which again makes physical sense.

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