The de Brogile wavelength associated with an electron accelerated through a potential difference of 100 V will be:

0.123 nm

The correct answer is Option (2) → 0.123 nm

Given,

h = Planck's constant = $6.626×10^{-34}Js$

m = mass of electron = $9.11×10^{-31}kg$

e = charge of electron = $1.6×10^{-19}C$

V = potential difference = 100 V

de Broglie wavelength = $λ=\frac{h}{\sqrt{2meV}}$

$=\frac{6.626×10^{-34}}{2×9.11×1.6×10^{-48}}$

$=\frac{6.626×10^{-34}}{5.399×10^{-24}}$

$=1.227×10^{-10}$

$=0.123nm$