Question

What is the de broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1 kev? 173 views?

Answer 1

heya your answer is here...

potential difference (V)= 1kev
$de \: broglie \: wavelength = \frac{12.27}{ \sqrt{v} } nm \\ = \frac{12.27}{ \sqrt{1 } } \\ = 12.27nm$