Recent content by genloz

ok.. so high energy muons won't stop, but why aren't low energy muons useful?

Hi, I'm starting a prac on muon lifetimes and we have been asked to calculate the range of energies suitable for the experiment... Thinking about this, I was a bit stumped as to what we were supposed to be calculating... Is this a limitation of the detection equipment? Or do low energy...

Okay, I think I get it.. the mass is only 938.8 or 939.7 because only 1 nucleon is excited, yeah? and the E is the actual value of the excitation, not the difference between excitation and ground state... so that: \Delta x = \frac{197 MeV fm}{\sqrt{2*938.8*3.055}} \Delta x = 2.6fm Which is still...

I guess the width should be hbar/meanlife but I don't think that matters in this instance...

pi- I:1 I3:1 Tz: ?? pi0 I:1 I3:0 Tz: ?? n I:1/2 I3:1/2 Tz:uud 1/2-1/2-1/2:-1/2 K- ubar s I: 1/2+0:1/2 I3:-1/2+0: -1/2 Tz: ?? Sigma- dds I:1/2+1/2+0:1 I3:-1/2+-1/2+0:-1 Tz: -1/2-1/2-1/2 Tz: u c t neutrinos 1/2 Tz: d s b charged leptons -1/2So...

Okay, so if the base level is 870keV, then the total excitation energy is 3055+870=3925... If we say 17u = 6*mp+8*mn+6*me then we arrive at: \Delta x = \frac{197 MeV fm}{\sqrt{2*(6*938.8+8*939.7+6*0.511)*3.925}} \Delta x = \frac{197 MeV fm}{\sqrt{2*(6*938.8MeV+8*939.7MeV+6*0.511MeV)*3.925MeV}}...

Well, for: ^{17}O (oxygen) r=r_{0}A^{1/3} r=1.3*17^{1/3}=3.34 and using the other way: \Delta x=\frac{\hbar}{\sqrt{2mE}} Knowing that the first excitation level of 17-oxygen is 3055keV \Delta x=\frac{0.197}{\sqrt{2*17*3055}} \Delta x=6.11*10^{-4} units?? m I guess? Clearly not the...

Homework Statement I had a whole list of reactions and had to show whether or not they could occur... I understood most of the reasons, bar the following, but was more confused about whether or not a reaction was strong or weak... I know that if the reaction involves a lepton its weak, and if...

okay, thanks... so a trick question? An even number of bosons would be required to conserve charge, but energy will never be conserved so it can't happen...

Is it possible to use Heisenberg again? eg. \Delta x \Delta p \approx \hbar \Delta x \approx \frac{\hbar}{\sqrt{2mE}} so subbing in the excitation level for E and the A value for m?

Okay, process 3 also isn't possible due to the delta having spin 3/2, the pion having spin 0 and the neutron having spin 1/2 so this is also impossible due to spin not being conserved? Process 4 Baryon Number 0 + 1 -> 1 + 0, conserved Spin 0 + 1/2 -> 1/2 + 0, conserved So process 4 is...

Actually, I think processes 3,4 and 5 are strong interactions rather than weak...

thanks very much! so D+ has c-dbar... D+ = 1869MeV D- = 1865MeV pi+ = 139.6MeV So that means both a W+ and a W- boson are required to keep the charge at +1 on both sides? And have energy greater than 139.6-5?

Ok, thanks very much... so the formula U = 1.44 (Z1Z/r) MeV fm already takes the coulomb force into consideration?

Oh, you're right.. sorry, so: r=2.3fm... That's still smaller than r=r0A^(1/3)=1.3*12^(1/3)=2.9762 (radius of the atom)... so that means it's closer than the electrons to the nucleus? Also what happened to coulomb repulsion?