- #1
Von Neumann
- 101
- 4
Problem:
a. Calculate the energy in eV of an electron with a wavelength of 1 fm.
b. Make the same calculation for a neutron.
Solution (so far):
a. λ=h/p=(hc)/(pc)=(1240 MeV fm)/(pc)=1fm
so, pc=1240 MeV
E=√[(pc)^2+E_0^2]
=√[(1240 MeV)^2+(.511MeV)^2]
∴E=1.24 GeV
This is the same answer as the back of my book, so I'm assuming this is the correct method of solution. However, I do the same thing for the neutron and my answer does not agree.
b. E=√[(1240 MeV)^2+(940 MeV)^2]
∴E=1560 Mev
My book says the correct answer is 616 MeV.
I don't see how an energy like that is even possible. Solving the following for pc,
E^2=(pc)^2+E_0^2
pc=√[E^2-E_0^2]
When you plug in the "correct" answer of E=616 MeV you get,
pc=√[(616 MeV)^2-(940 MeV)^2]
You certainly cannot take a square root of a negative number and get a meaningful answer. Any suggestions?
a. Calculate the energy in eV of an electron with a wavelength of 1 fm.
b. Make the same calculation for a neutron.
Solution (so far):
a. λ=h/p=(hc)/(pc)=(1240 MeV fm)/(pc)=1fm
so, pc=1240 MeV
E=√[(pc)^2+E_0^2]
=√[(1240 MeV)^2+(.511MeV)^2]
∴E=1.24 GeV
This is the same answer as the back of my book, so I'm assuming this is the correct method of solution. However, I do the same thing for the neutron and my answer does not agree.
b. E=√[(1240 MeV)^2+(940 MeV)^2]
∴E=1560 Mev
My book says the correct answer is 616 MeV.
I don't see how an energy like that is even possible. Solving the following for pc,
E^2=(pc)^2+E_0^2
pc=√[E^2-E_0^2]
When you plug in the "correct" answer of E=616 MeV you get,
pc=√[(616 MeV)^2-(940 MeV)^2]
You certainly cannot take a square root of a negative number and get a meaningful answer. Any suggestions?