# De Broglie's wavelength, what is it?

Gold Member
From an exercise I calculated the de Broglie's wavelength of a bullet (mass=40g and speed =1000m/s) to be very small: $$1.65651 \times 10 ^{-35}m$$. What does this mean? That to observe some interference pattern if I throw up these kind of bullets into a double slit, I'd need a width of the slit of the order of approximately $$10^{-35}m$$ (which is I know of course totally impossible)?
I must imagine the bullet as being a gamma ray with extremely small wavelength if it were a photon... I just don't understand the physical meaning of the de Broglie's wavelength. I have "read" the wikipedia's article on it but it's still unclear to me. Time to sleep now, maybe tomorrow I'll understand better the article, it's too late now for me. Any help is welcome too! Thanks in advance.

## Answers and Replies

RedX
lol, according to most physics textbooks, you would observe interference with a bullet...because most physics textbooks talk about infinitismal slits when talking about double slits! So no matter what the wavelength you should observe higher order beams when the two slits are separated by more than the bullet wavelength.

However, even if you could have a slit that small, would the bullet fit through it? How does quantum mechanics treat particles that aren't point particles?

Also, if you have no slits at all, does not the atomic spacing acts as slits? If atomic spacing is about an Angstrom (10^-10 ?) then any radiation with greater wavelength than that (such as light) should be diffracted no? So why do you need slits at all for photons?

Last edited:
PhilDSP
Hi fluidistic, I don't think the concept of a de Broglie wave for the entire bullet is a valid or realistic one. De Broglie described his equations as relating a particle to an accompanied wave. The macroscopic bullet is actually very nearly empty space because of the very great relative distances between the atoms or molecules. Empty space even if it is somehow deemed to be moving doesn't generate a wave whereas a particle under his theory can and does.

A better picturization might be a conglomeration of particles which each have a separate wave associated. Then you end up with a grid of waves.

Gold Member
Thank you guys. I think I understand the meaning of de Broglie's wavelength now.
To observe any wavelike property of the bullet, it should be reduced to approximately $$10^{-35} m$$ and matter too should be reduced to this totally impossible lengths. By matter I mean the distance between atoms for example. In this case I'd observe some diffraction of the bullets. But as you both said, it's not realistic, it's even impossible so it doesn't really make sense to talk about the wavelength of a macroscopic object such as a bullet.
Thank you. Oudeis Eimi
The macroscopic bullet is actually very nearly empty space because of the very great relative distances between the atoms or molecules.

Uh? That may be true for a gas, but not for a condensed lump of matter.

lightarrow
lol, according to most physics textbooks, you would observe interference with a bullet...because most physics textbooks talk about infinitismal slits when talking about double slits! So no matter what the wavelength you should observe higher order beams when the two slits are separated by more than the bullet wavelength.

However, even if you could have a slit that small, would the bullet fit through it? How does quantum mechanics treat particles that aren't point particles?

Also, if you have no slits at all, does not the atomic spacing acts as slits? If atomic spacing is about an Angstrom (10^-10 ?) then any radiation with greater wavelength than that (such as light) should be diffracted no? So why do you need slits at all for photons?
Do you think it's correct to say that the slits should act as regions of no interaction between the screen and the radiation? Then you couldn't use atomic spacing with visible em radiation because such radiation does interact with the atoms. X-rays instead interact less (with the outer electrons) so in this case atomic spacing really acts as diffraction grating.

With shorter spacings, its more and more difficult to find something that doesn't interact, because the fields (electromagnetic or nuclear) extends in those regions.

RedX
Do you think it's correct to say that the slits should act as regions of no interaction between the screen and the radiation? Then you couldn't use atomic spacing with visible em radiation because such radiation does interact with the atoms. X-rays instead interact less (with the outer electrons) so in this case atomic spacing really acts as diffraction grating.

With shorter spacings, its more and more difficult to find something that doesn't interact, because the fields (electromagnetic or nuclear) extends in those regions.

Well if you have a screen that is opaque to the radiation, and cut holes in it, then some radiation can leak through. If you now replace the gaps in the screen with the same material as the rest of the screen, then radiation cannot leak through. Therefore it must be true that the field created by the gaps (if the gaps were replaced by material) plus the field created by the screen with holes in it are equal and opposite. So because this is true, you can consider just the gap material, since this will give you the same result as a screen with holes in it (with the sources all in the screen): the two different setups differ by a negative sign which doesn't affect interference. So this is my understanding as to why for diffraction you can assume there are sources at the holes and nowhere else.

The atomic spacing is smaller than the wavelength of light (I think). Therefore it seems there should be diffraction even if the screen has no slits, since real screens are not continuum but are atoms that have an atomic spacing.

Obviously I'm wrong, but I can't quite figure out where. Are the wavelength of X-rays larger than the atomic spacing? I'm pretty sure the wavelength of light is larger than the atomic spacing.

As for interactions with the electrons, it is true that the electrons in atoms are at the optical frequency (spectral lines of atoms). So if you consider some really long wavelengths like radio waves, would you observe diffraction in an opaque screen? Or what about a screen with slits say a centimeter thick? Or does the material just absorb the radio waves?

Gold Member
The atomic spacing is smaller than the wavelength of light (I think).
Right. My memory says around $$10^{-10}m$$ for atomic spacing and around $$10^{-7}m$$ for visible light (violet is 400 nm and red is 650/700 nm).
Therefore it seems there should be diffraction even if the screen has no slits, since real screens are not continuum but are atoms that have an atomic spacing.
To me it's like you have a slit with a much thiner width than the wavelength and hence it's like the case of a radio wave passing through a slit of some centimeters width.
If I'm right to think it this way, then I suggest you to see the thread https://www.physicsforums.com/showthread.php?t=457440.
Obviously I'm wrong, but I can't quite figure out where. Are the wavelength of X-rays larger than the atomic spacing? I'm pretty sure the wavelength of light is larger than the atomic spacing.
X-rays wavelength is around the same order of magnitude than atomic spacing, namely 1 angstrom or $$10^{-10}m$$.
[/QUOTE]
I can't help for the rest of your interesting questions, I'm just a student.

RedX
Right. My memory says around $$10^{-10}m$$ for atomic spacing and around $$10^{-7}m$$ for visible light (violet is 400 nm and red is 650/700 nm).

Yeah I remember that 10-10 is the size of the atom. I just wasn't sure if the spacing between atoms was smaller or bigger than the size of an atom.

To me it's like you have a slit with a much thiner width than the wavelength and hence it's like the case of a radio wave passing through a slit of some centimeters width.
If I'm right to think it this way, then I suggest you to see the thread https://www.physicsforums.com/showthread.php?t=457440.

I actually think that would be true. Radiowaves passing through a slit of centimeters width
should spready out everywhere on the screen, just like this picture of water waves:

http://www.flickr.com/photos/valentinus/3341516456/ [Broken]

Last edited by a moderator:
eaglelake
Bullets are not quantum particles and we always generate a lot of confusion when we insist that they are. Assuming that there are no external forces acting, a bullet that goes through a slit travels in a straight line, as mandated by Newton’s first law. There isn’t much more we can say about this!

So, I will try to answer your question, “deBroglie’s wavelength, what is it?” I will give you the definition from quantum mechanics. If we perform a quantum experiment where we measure the momentum of a free particle, then we must expand the statefunction in terms of momentum eigenfunctions $$\varphi (\vec k,\vec r) = e^{i\vec k \cdot \vec r}$$ , where $$k = \frac{{2\pi }}{\lambda } = \frac{p}{\hbar }$$. We see that the deBroglie wavelength $$\lambda$$ is the wavelength of the momentum eigenfunction corresponding to the momentum eigenvalue $$p = \frac{h}{\lambda }$$.
Best wishes

lightarrow
Bullets are not quantum particles and we always generate a lot of confusion when we insist that they are. Assuming that there are no external forces acting, a bullet that goes through a slit travels in a straight line, as mandated by Newton’s first law. There isn’t much more we can say about this!
Where do you put the boundary quantum/classical? A bullet of a nanogram of lead is quantum or classical?

lightarrow
Well if you have a screen that is opaque to the radiation, and cut holes in it, then some radiation can leak through. If you now replace the gaps in the screen with the same material as the rest of the screen, then radiation cannot leak through.
Which radiation? EM radiation with frequency greater than that of X rays does leak through.

eaglelake
Where do you put the boundary quantum/classical? A bullet of a nanogram of lead is quantum or classical?

There is no boundary that clearly distinguishes a classical system from a quantum one. Besides, it is not the actual “size” of the particle, but the entire experimental apparatus that determines whether we use quantum mechanics or classical physics. For example, an electron in a cathode ray tube behaves as a classical particle. No quantum mechanics is required. But, an atomic electron violates the laws of classical mechanics and quantum mechanics is necessary. Actually, we should say that the entire atom, not the electron alone, is a quantum system.

A nanogram of lead isn’t anything until you do an experiment with it. If you fire it through a very wide slit, i.e. the slit width is much greater than its deBroglie wavelength, then it will travel in a straight line just like the bullet does. If you replace the wide slit with a very narrow one, then it can be deflected at different angles and you might see interference effects. Obviously, the narrow slit experiment is quantum mechanical. So, it is not the nanogram of lead by itself that is classical or quantum. Rather, it is what you decide to do with it.
Best wishes