# Homework Help: Physical meaning of a result in Bragg's diffraction

1. Sep 21, 2011

### fluidistic

1. The problem statement, all variables and given/known data
The distance between crystal planes in a KCl crystal is about $3.14 \times 10 ^{-10}m$. Calculate the Bragg's reflexion angle of first order for electrons with kinetic energy of 4keV. Compare it with photons that have the same energy.

2. Relevant equations
$\lambda n =2d \sin \theta$.
For an electron, $E _K =(\gamma -1 ) m_e c^2$.
$\lambda _B =\frac{h}{p}$ where $p=\gamma m_e v$.
For photons, $E=\frac{hc}{\lambda}$.
3. The attempt at a solution
Using the equations above and after some algebra, I calculated the velocity of the electrons to be $21530815.57 \frac{m}{s}$, somehow comparable to c.
In the end I found out that for electrons, $\sin \theta \approx 0.053378361$. I am not sure if theta is in degree or radian. I considered degrees so this gave me about $3.059809004°$.
I've done the same for photons (I noticed that they are X-rays since their wavelength is of the order of the Angstrom). I reached $\theta =29.59784517°$.

I'd like to understand why the results are different and why are the electrons less deviated by the atoms in the crystal than the photons with the same energy.
I guess the mass, the charge and the spin play a role, but I'd like to know which one is greater, etc.
If you have any comment(s) on this, please do it. I'm eager to learn.
Thank you.

2. Sep 21, 2011

### rl.bhat

For the same energy, have you compared the wave lengths of photon and electron?

3. Sep 21, 2011

### fluidistic

Hi.
Yes I have. For the electron: around 3.35 x10^-11 m.
For the photon: around 3.10x10^-10m.
So it's obvious that the photon will suffer a bigger angle of diffraction (I have in mind the undergrad lab experiment of diffraction patterns, violet/blue vs red light. The red light, larger wavelength was "more" diffracted than any other color in the visible).
But I don't understand why it is so. For the same energy, a photon will have a larger wavelength than a massive particle; I don't understand why.

4. Sep 22, 2011

### rl.bhat

momentum p = (2mE)^1/2
For photon E = hc/λ or λp = hc/E
For electron λe = h/p = h/ (2mE)^1/2
so λp/λe = c(2m/E)^1/2
Substitute the values and find the ratio.
As you can see the difference in the wave length is due to mass of the electron only.

Last edited: Sep 22, 2011
5. Sep 22, 2011

### ehild

The energy of the electron is equal to its KE + the energy of the rest mass E=KE+mec2. If the photon has this energy, its wavelength is much shorter than that of the electron. I guess you took the photon energy 4KeV.

ehild

6. Sep 22, 2011

### fluidistic

Strangely to me, I reach a ratio of 9.23x10^-4 so almost 10^-3. This implies that the electron has a larger wavelength than the photon (by at least a factor 1000), something that goes in counter of my previous result.
If I'm not diffracted with visible object, isn't it because my de Broglie's wavelength is very, very, very short compared to the size of the visible object?
I had never seen the momentum expressed under the form p=(2mE)^1/2. Also I think I should take the relativistic momentum of the electron like I did, because of it's speed.
Hmm I see. Indeed, I took a photon with E=4keV.

7. Sep 22, 2011

### ehild

p=(2mE)1/2 is the classical relation between the momentum and kinetic energy.
I think the problem should be understood that the photon has the same energy as the KE of the electron, 4 KeV. So your solution was correct.
As the numeric data were not given with more than 3 significant digits, do not give your results with more digits. v/c <0.1, ignoring its square means about 1 % error, so you can use non-relativistic approximation.

You can also get the momentum of the electron using the relation between total energy E=KE+mec2 and momentum p:

E2=p2c2+me2c4.

ehild