The problem statement:

Narrow beam of electrons hits an aluminium plate under an angle $\alpha=30^\circ$ according to the flat surface. We know the distance between two crystal planes in aluminium is $d=0.2nm$. At some voltage $U_1$ at wich electrons are accelerated we get a "mirror reflection" of electrons. If we increase the voltage to $U_2 = 2.25U_1$ we get same "mirror reflection" in the same direction. Calculate $U_1$.

What I have managed to do so far:

First of all I had to draw this sketch and i found out that in order to get the "mirror reflection" $\alpha=\beta$ crystal planes have to be parallel to the surface plane. So i concluded that $\alpha = \vartheta$. In any other way it would happen that $\alpha \neq \beta$ i think.

I hope my conclusion is correct and that $\vartheta = 30^\circ$.

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What i haven't managed to figure out:

Now I can use the Brag's law to calculate $N\lambda_1$ and $\underbrace{(N+1)}_{\llap{\text{I hope this is correct,}}\rlap{\!\!\!\text{but please someone confirm}}}\lambda_2$ where I don't know the number $N$.

\begin{align} \left. \begin{aligned} 2d\sin\vartheta &= N\lambda_1 \\ 2d\sin\vartheta &= (N+1)\lambda_2 \end{aligned}\quad \right\} \quad N\lambda_1 &= (N+1)\lambda_2\\ \frac{\lambda_1}{\lambda_2} &= \frac{N+1}{N}\\ \frac{\lambda_1}{\lambda_2} &= 1 + \frac{1}{N}\\ \frac{1}{N} &= \frac{\lambda_1}{\lambda_2} - 1\\ N &= \frac{1}{\lambda_1/\lambda_2 - 1} \end{align}

At this point I can see that I can calculate $N$ if I can find a ratio $\lambda_1/\lambda_2$. Does enyone have any idea how to find it?


I have been trying to get the ratio using the invariant interval, but it didn't work out for me. I can't get the ratio because factor $2.25$ is giving me trouble... Because i can't factor it out I can't get rid of variable $U_1$:

\begin{align} \frac{\lambda_1}{\lambda_2} &= \frac{h p_2}{p_1 h} = \frac{p_2}{p_1} = \frac{\sqrt{ { {E_{k2}}^2 + 2E_{k2}E_0} }\cdot c }{\sqrt{ {E_{k1}}^2 + 2E_{k1}E_0 }\cdot c} = \sqrt{ \frac{ e^2{U_{2}}^2 + 2eU_{2}E_0}{ e^2{U_{1}}^2 + 2eU_{1}E_0}} =\\ &= \sqrt{\frac{2.25^2\, e^2 {U_{1}}^2 + 2.25\, 2eU_{1}E_0}{e^2{U_{1}}^2 + 2eU_{1}E_0}} \end{align}


1 Answer 1


The energies involved should be small compared to the rest mass of the electron, so you can use Newtonian mechanics to get the momentum. We have $p_1=\sqrt {2mU_1},p_2=\sqrt {2mU_2}$, so $\frac {\lambda_1}{\lambda_2}=\frac {p_2}{p_1}=\sqrt{\frac{U_2}{U_1}}=1.5$

  • $\begingroup$ I like this but is there a correct way to do it relativistically. $\endgroup$
    – 71GA
    Jul 25, 2013 at 18:41
  • $\begingroup$ I always have a problem, because i totaly forget that we can deal with a problem in a nonrelativistic aproximation. But if a relativistic derivation is possible it would be really nice. $\endgroup$
    – 71GA
    Jul 25, 2013 at 18:49
  • $\begingroup$ The problem poser chose 2.25 because it came out nicely. I don't think there is a selection in the relativistic case that works cleanly. You would probably be left with a numeric solution instead of algebraic. $\endgroup$ Jul 25, 2013 at 19:06
  • $\begingroup$ It really works fabulous if nonrelativistic approach is used. Can you just tell me how, did you know that nonrelativistic should be used here? When did you first think of this? What was the trigger... ? $\endgroup$
    – 71GA
    Jul 25, 2013 at 21:24
  • $\begingroup$ Just that scattering like this is done at X-ray energies, while the electron rest mass is 0.5 MeV. You can check that the energies you find are well below that. $\endgroup$ Jul 25, 2013 at 21:28

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