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Chapter 44: Q7P (page 1363)

The rest energy of many short-lived particles cannot be measured directly but must be inferred from the measured momenta and known rest energies of the decay products. Consider the 蚁0meson, which decays by the reaction蚁0鈫�蟺++蟺-. Calculate the rest energy of the 蚁0meson given that the oppositely directed momenta of the created pions each have a magnitude 358.3 MeV. See Table 44-4 for the rest energies of the pions.

Short Answer

Expert verified

The rest energy of the rho meson is 769.06 MeV.

Step by step solution

01

Given data

Momenta of the outgoing pions

p=358.3MeVc

Rest mass of the pion

m0=139.6MeVc2

02

Determine the relativistic energy

The total relativistic energy of a particle of momenta and rest mass is

E=p2c2+m02c4 ..... (I)

03

Determining the rest energy of the rho meson

From equation (I), the total relativistic energy of each pion is

E=358.3MeVc22脳c2+139.6MeVc22脳c4=128378.89MeV2+19488.16MeV2=384.53MeV

From energy conservation, the rest energy of the rho meson is as follows:

E0= 2 x E

Substitute the values and solve as:

E0=2脳384.53MeV=769.06MeV

Thus, the required energy is 769.06 MeV.

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