Decay including massless partners

Problem:

A particle of rest mass M disintegrates at rest into a particle of rest mass m1, a particle of rest mass m2, and a high-energy photon (a gamma ray).  Find the maximum possible energy of the gamma ray in the rest frame of the initial particle of rest mass M.

Solution:

Problem:

A π0 meson decays into two γ-rays while at rest or in flight:  π0 --> γ + γ.

image

(a)  If the decaying π0 has velocity v and rest mass mπ , and the γ-ray is emitted at an angle θ with respect to the original direction of the π0, find the γ-ray energy as a function of mπ, v, and θ.
(b)  What is the maximum and minimum energy an emitted γ-ray can have, and at what emission angles do these occur?

Solution:

Problem:

A beam of π+ particles (mπ+ = 140 MeV/c2) is accelerated, so it has momentum pπ+ = 2 GeV/c in the laboratory frame.  A π+ particle decays into a positive muon (mμ = 105 MeV/c2) and a muon neutrino (mνμ = 0 MeV/c2) . 
(a)  What is the energy of the muon in the rest frame of the π+ particles?
(b)  What is the energy of the muon in the laboratory frame if it is found traveling in the beam direction?
(c)  What is the energy of the muon in the laboratory frame if it is found traveling at 0.25 rad with respect to the beam direction?

Solution:

Problem:

Cs55137 is a common laboratory radioactive source of electrons and gamma rays.  Eight percent of the time Cs55137 beta decays to the ground state of Ba56137.  The net atomic mass difference between the two isotopes is 1.18 MeV/c2.  A 180 degree spectrometer is used to measure the beta decay spectrum.  The spectrometer has a radius R = 3.8 cm.
image
(a)  Write down the reaction for the beta decay.
(b)  Calculate the maximum momentum of the beta decay electron/positron.  Express the result in MeV/c.
(c)  What is the vector direction of the spectrometer magnetic field relative to the drawing?
(d)  What is the magnetic field setting of the spectrometer for the maximum energy of the electron/positron to arrive at the detector?  Provide a numerical answer with units.

Solution:

Problem:

A free atomic nucleus at rest with mass M makes a transition from excited to ground state by emitting a γ-ray photon.
(a)  Find the energy of the γ ray photon and the kinetic energy of the recoil nucleus if the excitation energy was ΔE.  
(b)  Calculate the recoil energy of the 191Ir nucleus if the excitation energy ΔE = 129 keV.  

Solution:

Problem:

The question of the proton stability is one of the center questions of Grand Unification theories.  The present experimental limit of the proton lifetime from the Super Kamiokande experiment is τp > 1.6·1033 y.  More sensitive experiments have been proposed.  One of the possible proton decay modes is p --> e+ + π0, followed by the decay of the neutral pion into two photons: π0 --> γ + γ
Assume that the protons are at rest.
(a)  What is average flight distance of the neutral pion before its decay?
(b)  What are the maximum and minimum energies the photons can have?

Mp = 938.27 MeV
Mπ = 134.98 MeV
Me = 0.511 MeV
τπ = 8.4·10-17 s

Solution:

Problem:

A pion (π) decays at rest into a muon (μ) and a neutrino (ν).  In terms of the masses mπ and mμ (use the approximation mν = 0), find:
(a)  the momentum, energy, and velocity of the outgoing muon.
(b)  In the rest frame of the outgoing muon, what is the energy of the neutrino?  What was the initial velocity of the pion in this frame?

Solution:

Problem:

A pion (mass Mp) decays into a muon (mass Mμ) and a neutrino (massless).
(a)  Find the kinetic energy of the muon in the rest frame of the pion.
(b)  The muon is unstable, with a lifetime τ1 in the pion's rest frame.  Find the lifetime t0 of the muon in its own rest frame.

Solution:

Problem:

An excited nucleus of 57Fe formed by the radioactive decay of 57Co emits a gamma ray of 1.44 *104 eV.  In the process, there is conservation of energy and m0c2 = γma0c2 + hf,  where m0c2 is the initial mass of the nucleus and ma0c2 is its mass after the emission of the gamma ray.  There is also conservation of momentum, hf/c = γma0u, where u is the recoil velocity of the iron nucleus.  The energy released by the reaction is Er = (m0 - ma0)c2.
(a)  Show that  hf = Er(m0 + ma0)/(2m0) = (1 - Er/(2m0c2))Er.
Thus hf < Er: part of Er goes to the photon, and the other part supplies kinetic energy to the recoiling nucleus.

(b)  Set m0 = 57*1.7*10-27 kg, and show that Er/(2m0c2)) ~ 1.3*10-7.
Thus the fraction of the available energy Er that appears as recoil is small.

(c)  Mössbauer discovered in 1958 that, with solid iron, a significant fraction of the atoms recoil as if they were locked rigidly to the rest of the solid.  This is the Mössbauer effect.  If the sample has a mass of 1 gram, by what fraction is the gamma ray energy shifted in the recoil process?

(d)  A sample of normal 57Fe absorbs gamma rays of 14.4 keV by the inverse recoilless process much more strongly than it absorbs gamma rays of any nearby energy.  The excited nuclei thus formed reemit 14.4 keV radiation in random directions some time later.  This is resonant scattering.  If a sample of activated 57Fe moves in the direction of a sample of normal 57Fe, what must be the value of the velocity v that will shift the frequency of the gamma rays, as seen by the normal nuclei, by 3 parts in 1013?  This is one line width.

(e)  A Doppler shift in the gamma ray results in a much lower absorption by a nucleus if the shift is of the order of one line width or more.  What happens to the counting rate of a gamma-ray detector placed behind the sample of normal 57Fe when the source of activated Fe moves
    (i) toward the normal 57Fe,
    (ii) away from it?

(f)  If a 14.4 keV gamma ray travels 22.5 meters vertically upward, by what fraction will its energy decrease?
[Gravitation redshift, a thought experiment:  Suppose a particle of rest mass m is dropped from the top of a tower and falls freely with acceleration g.  It reaches the ground with a velocity v = (2gh)1/2, so its total energy E, as measured by an observer at the foot of the tower is E = mc2 + ½mv2 + O(v4) = mc2 + mgh + O(v4).
Suppose an observer has some magical method of converting all this energy into a photon of the same energy.  Upon its arrival at the top of the tower with energy E the photon is again magically changed into a particle of rest mass m' = E'/c. Energy conservation requires that m' = m. 
Therefore E'/E = mc2/(mc2 + mgh + O(v4)) = (1 + gh/c2 + O(v4))-1 = 1 - gh/c2 + O(v4).]

(g)  A normal 57Fe absorber located at this height must move in what direction and at what speed in order for resonant scattering to occur?

Solution: