How large is the second lattice spacing d₂ in graphite crystals?

The lattice spacing d₂ can be calculated with $$\begin{equation}d_\text 2=\frac{\frac{h}{\sqrt{2\cdot m_\text e \cdot e\cdot V_\text a}}}{2\cdot \sin\left(\frac{1}{2}\cdot \tan^{-1}\left(\frac{r}{L-R+\sqrt{R^2-r^2}}\right)\right)}\end{equation}$$So the acceleration voltage \(V_a\) and the corresponding radius \(r_{\rm{outer}}\) must be determined in the experiment:

$V_\text a$	6 kV	9 kV	12 kV
Radius router	1,65 cm	1,35 cm	1,15 cm

Using the measurements and the given values $m_\text e=9{.}1\cdot 10^{-31}\,\text{kg}$; $e=1{.}6\cdot 10^{-19}\,\text{C}$; $h=6{.}6\cdot 10^{-34}\, \text J \cdot \text s$; $d=2{.}13\cdot 10^{-10}\, \text m$; $L=12{.}7\,\text {cm}$; $R=6{.}35\,\text {cm}$, the lattice spacing d₂ in graphite can be calculated:

$V_\text a$	$$\lambda=\frac{h}{\sqrt{2\cdot m_{\rm e}\cdot e\cdot V_{\rm a} }}$$	Radius router	$$d_\text 2=\frac{\lambda_{\text{de Broglie}}}{2\cdot \sin\left(\frac{1}{2}\cdot \tan^{-1}\left(\frac{r}{L-R+\sqrt{R^2-r^2}}\right)\right)}$$
\(6\,\rm{kV}\)	$1.58\cdot 10^{-11} \,\text m$	\(1.65\,\rm{cm}\)	$1.20\cdot 10^{-10} \,\text m$
\(9\,\rm{kV}\)	$1.29\cdot 10^{-11} \,\text m$	\(1.35\,\rm{cm}\)	$1.20\cdot 10^{-10} \,\text m$
\(12\,\rm{kV}\)	$1.12\cdot 10^{-11} \,\text m$	\(1.15\,\rm{cm}\)	$1.23\cdot 10^{-10} \,\text m$
			Mittelwert $\left(1.21\pm 0.02\right)\cdot 10^{-10}\,\rm{m}$

Discussion of results

The average of the calculated lattice spacings $d_\text 2 = 1{.}21\cdot 10^{-10}\,\text{m}$ in graphite is close the the literature value $d_\text 2=1{.}23\cdot 10^{-10}\,\text m$ for the second lattice spacing. This is also supported by the low relative error. We can therefore safely conclude that the outer ring on the screen is a first order interference maximum from two lattice spacings $d_\text 2=1{.}23\cdot 10^{-10}\,\text m$.
This is confirmed by geometric considerations of the graphite crystal.

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