9. Considering the following three-body scattering process,...

9. Considering the following three-body scattering process, which is dominated by the following Feynman diagrams: \gamma \text{p} \Lambda^\ast K^+ K^- \gamma \text{p} \text{p} \phi K^+ K^- \text{p} Considering the above problem and five $\Lambda(\ast)$ hyperons with their masses (1116, 1405, 1520, 1670, 1690) MeV, draw the Dalitz plot of the process approximately again at several values of the total center-of-mass energy $\sqrt{s} \ge (M_{K^+} + M_{K^-} + M_p)$.

Transcript

00:01 We solved the given problem by first writing the generalized formulation reaction that is x is plus small x gives y, where x is a target nucleus and x is the incident light particle where y is the excited compound nucleus here in this case n e 20 we have here.

00:28 We assume x is initial.

00:30 At rest then by using a law of conservation of energy we can write m of x times c square plus m x c square plus k x kinetic energy is equal to m y c square plus k y c square plus k y plus e y where we have m x capital x and m small x and m capital y are the masses where kx and k y are the kinetic energies.

01:06 Ey here is the excitation energy of y.

01:11 Then by using a law of conservation of momentum we can write px is equal to p y.

01:18 Now we can write k y is equal to p y squared divided by two m y that is equal to p x squared divided by two m y that is this gives us mx divided by m y times kx.

01:40 So our k y is equal to m x divide by m y times kx.

01:46 Then we can write m x, mx times c squared plus, so this is capital x here, plus mx small x, c square, plus kx is equal to m y c square plus kx is equal to m y c square plus plus mx or my times kx plus ey.

02:15 From here we can solve for kx over kx will become my divided by my minus mx this times my minus m x minus m x m x x x times m x for part a of the problem, let small x represents the alpha particle and the capital x represents the oxygen 16 nucleus.

02:49 Then we can write mi minus mx minus m small x times c square.

02:57 We can write this by substituting values 19 .99244.

03:04 Atomic mass units minus 15 .99491 atomic mass units minus 4 .00260 units times the energy associated with one atomic mass unit which is 931 .5 mega electron volt per u.

03:30 So this gives us minus 4 .4 .5.

03:35 0 .722 mega electron volt.

03:39 Then we can find the kinetic energy of alpha particles.

03:44 So k alpha will be then 19 .99244 atomic mass unit divided by 19 .99244.

03:57 Atomic mass unit minus 4 .00260 atomic mass unit times minus 4 .00260 atomic mass unit times minus 4.

04:06 722 mega electron volt plus 25 mega electron volt this gives us approximately 25 .4 mega electron volt for part b of the problem, let's small x represents the proton and the capital x represents the f -19 fluorine nucleus then we can write mi m .y minus m capital x minus m small x times c square will be 19 .992 -44 minus 18 .99841.

04:52 This is in atomic mass unit, so i'm skipping the units here, minus 1 .00783.

05:00 This times 990841...

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