The wavelength of an electron of kinetic energy 4 . 50 × 10 - 29 J is ________ × 10 – 5 m . (Nearest integer) Given: mass of electron is 9 × 10 – 31 kg , h = 6 . 6 × 10 – 34 Js

Step-by-step solution

Given: Kinetic energy(K.E.)= 4 . 55 × 10 − 29 J Mass of electron ( m e ) = 9 . 1 × 10 − 31 kg h = 6 . 6 × 10 − 34 J − sec debroglie wavelength λ d = h mv = h 2 mKE Here λ 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 . λ = h 2 mKE = 6 . 6 × 10 - 34 2 × 9 × 10 - 31 × 4 . 5 × 10 - 29 = 6 . 6 × 10 - 34 9 &#21