Question

an electron beam is accelerated by a potential difference of 1000v . what is wavelength of the wave associated with the electron beam ? pls explain clearly

Answer 1

hey mate here is your answer

hope it helps u

Answer 2

Answer:

▪️λ = 1228 A°

Given:

▪️m = 9.11×10^-31 kg

▪️V = 10 kV = 10^4

Solution:

▪️λ = h / √(2meV)

▪️λ = 6.63×10^-34 / √(2 × 9.11×10^-31 × 1.6×10^-19 × 10^4)

▪️λ = 1.228×10^-13 m

▪️λ = 1228 A°

▪️Therefore, de Broglie's wavelength of electron is 1228 A°.

Hopes it help you❤️❤️