The wavelength of an electron of kinetic energy 4 . 50 × 10 ^ - 29...

Given:
Kinetic energy(K.E.)=$4 . 55 \times {10}^{- 29}J$
Mass of electron$( {m}_{e} ) = 9 . 1 \times {10}^{- 31}kgh = 6 . 6 \times {10}^{- 34}J - sec$

debroglie wavelength ${\lambda}_{d} = \dfrac{h}{mv} = \dfrac{h}{\sqrt{2mKE}}$

Here $\lambda$ is the de Broglie wavelength, $h$ is Planck's constant, $m$is the mass of the particle, $v$ is its velocity, and $p$ is the momentum of the particle, which is equal to $mv$.

$\lambda = \dfrac{h}{\sqrt{2mKE}}$

$= \dfrac{6 . 6 \times {10}^{- 34}}{\sqrt{2 \times 9 \times {10}^{- 31} \times 4 . 5 \times {10}^{- 29}}}$$= \dfrac{6 . 6 \times {10}^{- 34}}{9 \times {10}^{- 30}} = \dfrac{6 . 6}{9} \times {10}^{- 4}$

$= \dfrac{66}{9} \times {10}^{- 5}$

$= 7 . 33 \times {10}^{- 5}$