Is the outer circle a second order interference maxima?

If the outer circle is a second order interference maximum, it must satisfy: $$\frac{h}{\sqrt{2\cdot m_{\text e}\cdot e\cdot V_{\text a} }}=\lambda_{\text {de Broglie}}=d\cdot \sin\left(\frac{1}{2}\tan^{-1}\left(\frac{r}{L-R+\sqrt{R^2-r^2}}\right)\right)$$ To see if this is true, the radius of the outer circle on the screen of the electron diffraction tube must be determined for various acceleration voltages Ub. Here are three example measurements from the experiment:
$V_\text a$ 5 kV 7 kV 10 kV
Radius router 1.80 cm 1.55 cm 1.25 cm
Using these measurements and the given sizes $m_\text e=9{.}1\cdot 10^{-31}\,\text{kg}$; $e=1{.}6\cdot 10^{-19}\,\text{C}$; $h=6{.}6\cdot 10^{-34}\, \text J \cdot \text s$; $d=2{.}13\cdot 10^{-10}\, \text m$; $L=12{.}7\,\text {cm}$; $R=6{.}35\,\text {cm}$ the wavelengths can be calculated:
$V_\text a$ $$\lambda=\frac{h}{\sqrt{2\cdot m_{\rm e}\cdot e\cdot V_{\rm a} }}$$ Radius router $$\lambda_{\text {Ex}}=d\cdot \sin\left(\frac{1}{2}\tan^{-1}\left(\frac{r}{L-R+\sqrt{R^2-r^2}}\right)\right)$$ relative error
5 kV $1.73\cdot 10^{-11} \,\text m$ 1,80 cm $1.53\cdot 10^{-11} \,\text m$ 11.56 %
7 kV $1.46\cdot 10^{-11} \,\text m$ 1,55 cm $1.31\cdot 10^{-11} \,\text m$ 10.27 %
10 kV $1.22\cdot 10^{-11} \,\text m$ 1,25 cm $1.05\cdot 10^{-11} \,\text m$ 13.93 %

Discussion of results

The clearly higher relative deviation between the calculated wavelengths shows that the second circle is presumably not a second order interference maximum, but rather must be another first order interference maximum.
This is supported by the observation that the outer circle is equally intense (bright) as the inner circle. A second order diffraction maximum would be however less bright.
Graphite must have a second plane spacing \(d_2\) that can fulfill the Bragg condition and therefore can produce the outer interference maximum.

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