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kogti [31]
2 years ago
8

An electron, starting from rest, accelerates through a potential difference of 40.0 V. What is the final de Broglie wavelength o

f the electron, assuming that its final speed is not relativistic?
Physics
1 answer:
kondor19780726 [428]2 years ago
6 0

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

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