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

1.94 x 10^-10 m

Step-by-step explanation:

V = 40 V

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

`\lambda = (12.27)/(√(V)) Angstrom`

`\lambda = (12.27)/(√(40)) Angstrom`

λ = 1.94 Angstrom = 1.94 x 10^-10 m