Question

Through what potential difference must an electron be accelerated from rest to have a de Broglie wavelength of 500 nmnm

Answer 1

Answer:

V = 6.04 10⁻⁶ V

Explanation:

The energy in a system is conserved so the potential energy of the system must be transformed into the kinetic energy of the electron.

              $E_{p}$ = K

              eV = ½ m v²

              V = ½ m v²/e

Let's use the de Broglie relation

             λ= h / p

the moment is

            p = mv

let's replace

            λ = h / mv

            v = $\frac{h}{m \lambda }$

hence the potential energy

             V = ½ $\frac{m}{e}$ (\frac{h}{m \lambda })²

            $V = \frac{h^{2} }{2e m \lambda^{2} }$

let's reduce λ to sistem SI

             λ = 500 nm = 500 10⁻⁹ m = 5 10⁻⁷ m

let's calculate

             V = (6.63 10⁻³⁴)² / (9.1 10⁻³¹  1.6 10⁻¹⁹ (5 10⁻⁷)² )

         

             V = 6.04 10⁻⁶ V