Answer to Question #100137 in General Chemistry for Daisy Godoy

Question

an electron (me = 9.10939 × 10–28 g) moving at 3.85 × 106 m/s in an electron microscope

nm =?

Answer 1

$$\lambda = {h \over m_ev}$$

$$\lambda = {6,63∙10^{-34} \over 9.10939 ∙ 10^{–28}∙3.85 ∙ 10^6} = 0.18904∙ 10^{-12} m = 1.8904∙ 10^{-4} nm$$