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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?

Respuesta :

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

[tex]\lambda = \frac{12.27}{\sqrt{V}} Angstrom[/tex]

[tex]\lambda = \frac{12.27}{\sqrt{40}} Angstrom[/tex]

λ = 1.94 Angstrom = 1.94 x 10^-10 m

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