# De Broglie wavelengths and electron diffraction

1. Jun 14, 2010

### DGriffiths

I am interested to know why in any description of electron diffraction apparatus they seem to suggest that the electrons need accelerating up to 5kV (or at least several kV) to show the electron diffraction rings, this seems to give a de Broglie wavelength of around 1.2 x 10^-11m whereas the lattice spacing in a graphite target is of the order of 2 x 10^-10m

Now working back from this to get de Broglie wavelength similar to this lattice spacing gives electron momentum of 3.32 x 10^-24kgm/s, ke of 6.03 x 10^-18J and so we are talking 37.8eV electrons????

Why isn't a 50V supply enough then why the need for 5kV?

2. Jun 14, 2010

Hello DGriffiths.
De Broglie wavelength=h/mv and eV=0.5mv^2.
From the above, wavelength=h/(2eVm)^0.5
I am wondering if you used an incorrect equation or made a mistake in your calculations somewhere.

3. Jun 14, 2010

### dlgoff

Last edited by a moderator: Apr 25, 2017
4. Jun 15, 2010

### alxm

I don't think you can run a CRT at 50 V.

I suspect the 5kV isn't required for the diffraction itself, but to literally show the pattern.

5. Jun 15, 2010

### xepma

Perhaps diffraction will take place at those speeds, but this is hardly observable. To actually generate a nice diffraction pattern you need the ratio $$\lambda / d$$ to be small (and not of order 1). Here $$\lambda$$ is the de Broglie wavelength and d the lattice spacing.

The reason is that the peaks with minimum intensity are positioned at angles $$\theta_n$$ (as measured from the normal vector of the lattice)

$$\sin(\theta_n) = n\lambda / d$$

So if $$\lambda / d$$ is of order 1, i.e. the two are of equal size, then the first minimum peak is at 90 degrees -- so you won't see a diffraction pattern at all. You will need the wavelength to be, preferably, a factor of 10 smaller. That way you can at least spot a couple of peaks of the diffraction pattern.

6. Jun 15, 2010

### PhilDSP

Yes, and from what I've read about the G. P. Thomson experiments, at slower electron speeds the circular wave pattern was larger but at a certain point the intensity and subsequent resolution dropped too low to be usable.

7. Jun 15, 2010

### DGriffiths

Thanks for all the replies folks, I didn't think about the ability to see a nice diffraction pattern being dependent on the correct lambda/d but that makes sense to me so cheers