Maximum kinetic energy of a Pion produced via a kaon decay

[james walshe]

Homework Statement

A possible decay mode a of a positive Kaon in the production of three Pions as shown below:.
K+ → π+ + π+ + π−
Whats is the maximum Kinetic energy that anyone of the pions can have?
The kaon is a rest when it decays and so its momentum is 0.

Homework Equations

p(π+)=p(π−)=p
E(k)=E(π+) + E(π+) + E(π−)
Where E=√((pc)^2+(mc^2 )^2 )
(E^2)-(p)^2=(m)^2

The Attempt at a Solution

The mass of the koan:493.7Mev/c^2
The mass of a Pion: 139.6 Mev/c^2

AS The mass of the three Pions is the same the expression for the total energy becomes:
3(√((pc)^2+(m_pionc^2 )^2 )=m_kaonc²
and through rearranging I come up with a momentum of 80.2MeV.

Alternatively I tried to use the following approach:
E(pion)=1/3*(Ek)=164.7MeV
P=164.7Mev-139.6MeV=25.1Mev

However the book from which the question is taken from (Particle Physics By Anwar Kama: problem 3.19) gives the solution as 50Mev?

Any help on this would be greatly appreciated as it is driving me nuts trying to figure it out.

 

[Orodruin]

In general, the pions will not have the same momentum and thus not the same kinetic energy. The question is asking for the highest possible kinetic energy of a pion.

Edit: Also, given the precision of the given particle masses, the result could have been quoted with 3 or 4 significant digits: 48.04 MeV.

 

[james walshe]

If a particle decays at rest into three daughter nuclei of the same mass, what is the maximum kinetic energy of anyone of the daughter nuclei?

 

[ChrisVer]

The answer, without making any assumption for the 3 final body spatial momenta, is not a fixed number...it will depend on the momenta of the 2 of the 3 particles.

 

[mfb]

The answer, without making any assumption for the 3 final body spatial momenta, is not a fixed number...it will depend on the momenta of the 2 of the 3 particles.

The maximum is a fixed number.

@james walshe: If one pion gets the maximal energy, how will the decay look like? Can you say something about the directions the pions will fly to?

 

[Faija369]

Hello,
Firstly, in order to answer this question we will need the rest energies of all the particles in this equation with the help of a table representing
'particles and their properties', for instance, https://embed.gyazo.com/1b84502edbb60778ee4e66f21b4934b2.png

The equation of 'a K+ meson decaying into three charged mesons' is:
K+ → π+ + π+ + π−
(Charge is conserved)

Considering that the kaon is at rest before it decays we can use the table to find the rest energy of the Kaon, which would ultimately be the total energy of all the products including the pions and the kinetic energy; this is essential in order for all the energy to be conserved.
Henceforth,
494 Mev → (π+ + π+ + π−) + KE

Afterthat, I would substitute the rest energies of each pion referencing the same table,
494Mev → (140 MeV + 140 MeV + 140 MeV) + KE
Simplified:
494 Mev → 420 MeV + KE

Then, I shall subtract the total energy from the reactants (rest energy of the K+) from the total rest energy of the pions in order to have the total maximum kinetic energy of all the pions,
494 MeV - 420 MeV = 74 MeV

Therefore, 74 Mev is the maximum kinetic energy of all the pions, however, if you would like the specific total maximum energy of 'each' pion you shall then divide the total kinetic energy by 3 (assuming that they all have equal kinetic energy) which is,
74 MeV / 3 = 24.6666667 MeV of kinetic energy for each pion equally.

Hope I have helped (2 years later - heh),
Faija

 

[mfb]

(assuming that they all have equal kinetic energy)

They do not, as discussed 2 years ago already.

Giving full solutions is against the forum rules, but as this thread is two years old it does not matter any more.