I found two versions of the Diffusion coefficient, first:
$$D=\frac{\pi \lambda }{8}\overline{c}$$
Where $ \overline{c}$ ist the particles mean thermal velocity and $\lambda$ the particles mean free path. (Found in W. C. Hinds, Aerosol Technology. Wiley Interscience (1999). S.156)
and second the version from my lecture (also found on wikipedia):
$$D=\frac{ \lambda}{3} \langle v\rangle$$
Where $\langle v\rangle$ is the particles mean thermal verlocity.
Hinds gives two more equations (S.154):
$$c_{\text{rms}}=\sqrt{\frac{3kT}{m}}$$
$$\overline{c}=\sqrt{\frac{8kT}{\pi m}}$$
With $c_{\text{rms}}$ being the root mean square velocity of a particle, $k$ being the Boltzmann-constant, $T$ being the temperature and $m$ the particles mass. My first thought was that maybe there is a typo in the book or the lecture notes so I calculated
$$\frac{c_{\text{rms}}}{\overline{c}}=\sqrt{\frac{3\pi}{8}}$$
which led to
$$\overline{c}=\sqrt{\frac{8}{3\pi}}c_{\text{rms}}$$
so I replaced $\overline{c}$ in
$$D=\frac{\pi \lambda }{8}\overline{c}$$
just to come to
$$D=\frac{\pi \lambda }{8}\sqrt{\frac{8}{3\pi}}c_{\text{rms}}$$
Which (set $\langle v\rangle=c_{\text{rms}}$, due to the potential typo) because of the square root isn't
$$D=\frac{ \lambda}{3} \langle v\rangle$$
Where did I make a mistake?
