The de-Broglie wavelength associated with an electron, accelerated by a potential difference of 81 V is given by [2023]

Question

The de-Broglie wavelength associated with an electron, accelerated by a potential difference of 81 V is given by [2023]

Smart Achievers · Accepted Answer

(4)

The wavelength of electron using de-Broglie equation is, $λ_{e} = \frac{h}{m v} = \frac{h}{\sqrt{2 m e V}} Å$

$λ_{e} = \frac{12.27}{\sqrt{V}} Å = \frac{12.27}{\sqrt{81}} Å = \frac{12.27}{9} Å$

$= 1.36 Å (∵ 1 Å = 0.1 nm)$

$= 0.136$ $nm$

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