Question

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 ${\mathit{\rho }}^{\mathbf{0}}$meson, which decays by the reaction${\mathit{\rho }}^{\mathbf{0}}\mathbf{}\mathbf{\to }{\mathit{\pi }}^{\mathbf{+}}\mathbf{}\mathbf{+}{\mathit{\pi }}^{\mathbf{-}}$. Calculate the rest energy of the ${\mathit{\rho }}^{\mathbf{0}}$meson 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

The rest energy of the rho meson is 769.06 MeV.

Step-by-step solution

Given data

Momenta of the outgoing pions

$p=358.3\frac{\mathrm{MeV}}{\mathrm{c}}$

Rest mass of the pion

$m_0=139.6\frac{\mathrm{MeV}}{{\mathrm{c}}^2}$

Determine the relativistic energy

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

$\mathit{E}\mathbf{=}\sqrt{{\mathbf{p}}^{\mathbf{2}}{\mathbf{c}}^{\mathbf{2}}\mathbf{+}{\mathbf{m}}_{\mathbf{0}}^{\mathbf{2}}{\mathbf{c}}^{\mathbf{4}}}$ ..... (I)

Determining the rest energy of the rho meson

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

$\begin{array}{c}E=\sqrt{{\left(358.3\frac{\mathrm{MeV}}{{\mathrm{c}}^2}\right)}^2\times c^2+{\left(139.6\frac{\mathrm{MeV}}{{\mathrm{c}}^2}\right)}^2\times c^4}\\ =\sqrt{128378.89{\mathrm{MeV}}^2+19488.16{\mathrm{MeV}}^2}\\ =384.53\mathrm{MeV}\end{array}$

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

E₀= 2 x E

Substitute the values and solve as:

$\begin{array}{c}{E}_0=2\times 384.53\mathrm{MeV}\\ =769.06\mathrm{MeV}\end{array}$

Thus, the required energy is 769.06 MeV.