An electron is accelerated through a potential difference of 144 V. The de-broglie wavelength associated would be approximately:

0.1 nm

The correct answer is Option (3) → 0.1 nm

The kinetic energy is also related,

$K.E.=\frac{1}{2}mv^2$

$ev=\frac{1}{2}mv^2⇒v=\sqrt{\frac{2ev}{m}}$

$v=\sqrt{\frac{2ev}{m}}=\sqrt{\frac{2×1.6×10^{-19}×144}{9.11×10^{-31}}}$

$≃7.12×10^6$

Using the Momentum formula,

$P=mv=(7.12×10^6)×(9.11×10^{-31})=6.48×10^{-24}$

Wavelength, $λ=\frac{h}{P}=\frac{6.63×10^{-34}}{6.48×10^{-24}}$

$=0.1nm$