Question

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

Answer 1

V = 6.04 10⁻⁶ V

Step-by-step 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 =
`(h)/(m \lambda )`

hence the potential energy

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

`V = (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