$${\Delta L = \frac{4\alpha m}{\pi\rho D^2} = \frac{{4}{}\cdot {0,00018}{}\cdot {0,002}{}}{{3,14}{}\cdot(60\cdot{10^{-6}}{})^2\cdot{13,5\cdot 10^3}{}} \approx {9,4}\,\mathrm{mm}}$$