A proton in a linear accelerator has a de Broglie wavelength of 141 pm .

Question

A proton in a linear accelerator has a de Broglie wavelength of 141 pm .

What is the speed of the proton? (The mass of the proton is 1.673×10−271.673×10−27 kg and 1 J = 1 (kg⋅m^2)/s^2

Answer 1

Recall:

λ = h / mv

λ = wavelength in meters = 141 pm x 1 m / 1x10¹² pm = 1.41x10^-10 m

h = Plancks constant = 6.626x10^-34 Js

m = mass = 1.673x10^-27 kg

v = velocity = ?

1.41x10^-10 m = 6.626x10^-34 Js / (1.673x10^-27 kg) (v)

v = 6.626x10^-34 Js / (1.41x10^-10 m)(1.673x10^-27 kg)

v = 2809 Js / kg-m recall that J(joules) = kg*m²/s² and plugging this in for J in our answer, we end up with..

v = 2809 m/s