The de Brogile wavelength associated with an electron accelerated through a potential difference of 100 V will be:
Answer & explanation
Correct answer: option 2
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 = $λ=\frac{h}{\sqrt{2meV}}$
$=\frac{6.626×10^{-34}}{2×9.11×1.6×10^{-48}}$
$=\frac{6.626×10^{-34}}{5.399×10^{-24}}$
$=1.227×10^{-10}$
$=0.123nm$