Question

The subatomic particle called the Delta (1232) is an excited state ot the proton and has a life time of ~ 10-23 s. What is the uncertainty on its energy in eV? A proton (m = 938 MeV/c2) is confined in the volume of a nucleus 5 fm across. Assuming it moves only in one dimension, estimate the proton minimum kinetic energy. How would you change your result when considering a more realistic three dimensional nucleus?

Answer 1

a) The uncertainity relation between energy and time is ∆t*∆E ∼ h/4pi = 5.27*10^-35

Then   ∆E = 5.27*10^-35/10^-23 = 5.27*10^-12 J

b) The uncertainity relation between momentum and position is ∆x*∆p ∼ h/4pi = 5.27*10^-35

  Then   ∆p = 5.27*10^-35/5*10^-15 = 1.054*10^-20

Then minimum K.E . = (∆p)^2/2m = (1.054*10^-20 )^2/(2*1.67*10^-27) = 3.33*10^-14 J