The de Broglie wavelength of an electron having kinetic energy 56 eV is:

0.164 nm

The correct answer is Option (3) → 0.164 nm

Given,

K.E., Kinetic energy of an electron = 56eV

and,

$K.E.=\frac{P^2}{2m}$

P = momentum of $e^-$

m = mass of $e^-$

$P=\sqrt{2m\,K.E.}$

$=\sqrt{2×9.1×10^{-31}×56×1.602×10^{-19}}$

$=4.04×10^{-24}$

Now,

De-Broglie wavelength, $λ=\frac{h}{P}$

$λ=\frac{6.63×10^{-34}}{4.04×10^{-24}}$

$=1.64×10^{-10}m$

$=0.164nm$