What is the de Broglie wavelength associated with an elector accelerated through a potential difference of 144 V?

0.0102 m

The correct answer is Option (3) → 0.0102 nm

The De-Broglie wavelength (λ) associated with an electron is -

$λ=\frac{h}{\sqrt{2meV}}$

where,

e, charge of electron = $1.602×10^{-19}C$

V, accelrating potential difference = 144 V

$λ=\frac{h}{\sqrt{2×9.1×10^{-31}×2.3×10^{-17}}}=\frac{6.26×10^{-34}}{6.46×10^{-24}}$

$=0.0102nm$