Q. 3. Suppose one photon has an energy of 200 MeV and is traveling along the x axis. Suppose another has an energy of 100 MeV and is traveling along the y axis.
(a) What is the total energy of this system?
(b) What is the total momentum?
(c) If a single particle had this same total energy and momentum:
i. What would be its mass?
ii. In what direction would it be traveling?
iii. With what speed?
Solution:
- Given, E1 = 200 MeV, E2 = 100 MeV
Therefore,
Total energy, ETotal = E1+ E2 = (200+100) MeV = 300 MeV.
- For momentum calculation,
Energy of a photon, E=hc/λ = hc/(h/p), (λ=h/p)
Therefore, E = pc
For Particle having energy 200 Mev,
⇒ p1 = E/c
⇒ p1 = 200 x 106 x (1.6×10−19)/ (3× 108) kgm/s
⇒ p1 = 106.66 x 10-21 kgm/s
And, ⇒ p2 = 100 x 106 x (1.6×10−19)/ (3× 108) kgm/s
⇒ p2 = 53.33 x 10-21 kgm/s
Therefore, Total momentum,
P = p1 + p2
= 106.66 – 53.33
= 53.33 kgm/s.
(Note: Momentum is a vector quantity, so total momentum of the two photons is calculated by vectors addition rule of two vectors which are in opposite directions.)
- i. the photon which has mass when moving but a zero mass at rest
Therefore, the relativistic mass can be calculated by below formula,
E = m.c2
⇒ m =E/c2
= 200 MeV.
ii. 27o
iii. Speed(relativistic),
If the rest mass of a particle is m and the total energy is E , then
and,
Therefore,
Finally, speed, v = 0.74 c.
Q. 4. State which of the following processes can be observed and why. Are the corresponding charge conjugate processes observed? why?:
Solution:
(a) p + n p + μ+ + μ–
Nature has specific rules for particle interactions and decays, and these rules have been summarized in terms of conservation laws. One of the most important of these is the conservation of baryon number.
So, we now check Baryon number conservation,
Therefore, p + n p + μ+ + μ–
Charge, Q: 1 + 0 = 1 + 1 -1 (conserved)
Baryon no., B: 1 + 1 ≠ 1 + 0 + 0 (conserved)
From the above inequality we can conclude that this reaction is not allowed because non-conservation of baryon number.
(b)
Charge, Q: -1 = -1 (conserved)
Baryon number, B: 0 = 0 + 0 (conserved)
(c) n + P e–
Charge(Q): 0 + 1 ≠ -1 (Not conserved, therefore this reaction not allowed)
(d)
Q: -1 = -1 + 0 + 0 (conserved)
B: 0 = 0 + 0 + 0 (conserved)
L:1 = +1 -1 +1 (Conserved)
(e)
Q: 0 = 0 +0 (Conserved)
B: 1= 1 + 0 (Conserved)
Q. 5. The four particles are charmless, topless and bottomless baryons. They have a charge of Q = –1. The strangeness S, ordinary spin J and measured mass in MeV are given by the following numbers in brackets respectively:
(a) Find the quark composition of the four particles.
(b) How can the quark structure of Ώ– be reconciled with Pauli’s exclusion principle?
(c) How does the particle differ from the particle in a way which might explain qualitatively why their masses are different?
Solution:
- Δ– = ddd
= sdd
= dss
Ώ– = sss
The quark model and the exclusion principle were reconciled as a result of ideas developed by Oscar W. Greenberg of the University of Maryland at College Park and, independently, by Moo-Young Han of Duke University and Yoichiro Nambu of the University of Chicago. What is needed is to assume that each kind of quark can exist in any of three states. For example, if an s quark in state A is combined with an s quark in state B and an s quark in state C to form the omega-minus particle, the exclusion principle is saved (Rebbi, 1983).
The Xi baryons or cascade particles are a family of subatomic hadron particles which have the symbol Ξ and may have an electric charge (Q) of +2 e, +1 e, 0, or −1 e, where e is the elementary charge. Like all conventional baryons, they contain three quarks. Xi baryons, in particular, contain one up or down quark plus two more massive quarks: either strange, charm or bottom. They are historically called the cascade particles because of their unstable state; they decay rapidly into lighter particles through a chain of decays.
The Sigma baryons are a family of subatomic hadron particles which have two quarks from the first flavour generation (up and/or down quarks), and a third quark from higher flavour generations, in a combination where the wavefunction does not swap sign when any two quark flavours are swapped. They are thus baryons, with total Isospin of 1, and can either be neutral or have an elementary charge of +2, +1, 0, or −1. They are closely related to the Lambda baryons, which differ only in the wavefunction’s behavior upon flavour exchange
Reference:
[1] Rebbi, C. (1983). The Lattice Theory of Quark Confinement. Scientific American, 248(2), 54–65. https://doi.org/10.1038/scientificamerican0283-54
[2] Warfreak. (2010, September 29). Physics Forums: Science Articles, Homework Help, Discussion. Retrieved from https://www.physicsforums.com/threads/photon-energy-momentum-and-velocity.433429/.
[3]Physics Stack Exchange, Relativistic speed/energy relation, https://physics.stackexchange.com/questions/716/relativistic-speed-energy-relation-is-this-correct.
