The de-Broglie wavelength associated with an electron accelerated through a potential difference of 100 V is:

1.23 Å

The correct answer is Option (2) → 1.23 Å

The De-Broglie Wavelength (λ),

$λ=\frac{h}{P}=\frac{h}{\sqrt{2meV}}$

$⇒λ=\frac{6.63×10^{-34}}{\sqrt{2×9.11×10^{-31}×1.6×10^{-19}×100}}$

$=\frac{6.63×10^{-34}}{\sqrt{29.152}×10^{-50}}$

$=\frac{6.63}{5.39}×10^{-10}=1.23Å$