# Problem similar to Davison & Germer eksperiment but with more angles and the mirror reflection

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 $\vartheta = 30^\circ$.

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?

EDIT 1:

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}

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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 $\frac {\lambda_1}{\lambda_2}=\frac {p_2}{p_1}=\sqrt{\frac{U_2}{U_1}}=1.5$