What is the de Broglie wavelength of a 240-MeV proton?

Question

Hint: Use relativistic mechanics

Help...

Answer 1

Let L and p be de Broglie wavelength and the (relativistic) momentum. Let's give two ways to obtain L. The formula obtained by the first method is probably neater...

FIRST APPROACH (without computing the speed v)
The energy-momentum 4-vector is (E/c,p) = (E/c,h/L)
For a particle of rest mass M, the Minkowski square of the above is:
-(Mc)^2 = -(E/c)^2 + (h/L)^2
So (hc/L)^2 = E^2 - (Mc^2)^2 = (E-Mc^2)(E+Mc^2)
L = hc / sqrt ( (E-Mc^2)(E+Mc^2) )

The rest energy of a proton is Mc^2 = 938.27 MeV
For a 240 Mev proton
E - Mc^2 = 240 Mev
E+ Mc^2 = 240 + 2 Mc^2 = 2116.5 MeV
sqrt [(E-Mc^2)(E+Mc^2)] = 712.72 MeV = 1.1419 e-10 J

Recall that hc = 1.9864456 e-25 J.m

Therefore, the de Broglie wavelength L is given by
L = hc / (1.1419 e-10) = 1.73959 e-15 m
L = 1.74 fm (fm = femtometer = Fermi = 10^(-15) m )


SECOND APPROACH (featuring a computation of v)
The relation L = h/p is true with p = mv / sqrt(1-v^2/c^2)
So, L = (h/mv) sqrt(1-v^2/c^2)

The last factor is the ratio of the rest mass to the energy:
sqrt (1 - v^2/c^2 ) = 938.27 / (938.27 + 240)
Therefore:
v^2/c^2 = 1 - 1/(1+240/938.27)^2 = 0.3656
v = 0.6049 c = 1.8134 10^8 m/s
h/mv = 2.1846 fm
L = (h/mv) / (1 + 240 / 938.27) = 1.73959 fm
L = 1.74 fm
Same result, of course.