Question

An electron, starting from rest, accelerates through a potential difference of 40.0 V. What is the final de Broglie wavelength of the electron, assuming that its final speed is not relativistic?

Answer 1

Answer:

1.94 x 10^-10 m

Explanation:

V = 40 V

The relation for the de broglie wavelength and teh potential difference is given by

$\lambda = \frac{12.27}{\sqrt{V}} Angstrom$

$\lambda = \frac{12.27}{\sqrt{40}} Angstrom$

λ = 1.94 Angstrom = 1.94 x 10^-10 m