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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 $ρ^{0}$ meson, which decays by the reaction $ρ^{0} \to π^{+} + π^{-}$ . Calculate the rest energy of the $ρ^{0}$ 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

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

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

Rest mass of the pion

$m_{0} = 139.6 \frac{MeV}{c^{2}}$

02

Determine the relativistic energy

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

$E = \sqrt{p^{2} c^{2} + m_{0}^{2} c^{4}}$ ..... (I)

03

Determining the rest energy of the rho meson

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

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

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

E₀= 2 x E

Substitute the values and solve as:

$\begin{matrix} E_{0} = 2 \times 384.53 MeV \\ = 769.06 MeV \end{matrix}$

Thus, the required energy is 769.06 MeV.

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Most popular questions from this chapter

Question: Given figure shows the paths of two particles circling in a uniform magnetic field. The particles have the same magnitude of the charge but opposite signs. (a)Which path corresponds to the more massive particle? (b)If the magnetic field is directed into the plane of the page, is the more massive particle positively or negatively charged?

Question: An electron cannot decay into two neutrinos. Which of the following conservation laws would be violated if it did: (a) energy, (b) angular momentum, (c) charge, (d) lepton number, (e) linear momentum, (f) baryon number?

The $A_{2}^{+}$ particle and its products decay according to the scheme

$A_{2}^{+} \to ρ^{0} + π^{+} ρ^{0} \to π^{+} + π^{-} π^{+} \to μ^{+} + ν μ^{+} \to e^{+} + ν + \bar{ν} π^{-} \to μ^{-} + \bar{ν} μ^{-} \to e^{-} + ν + \bar{ν}$

(a) What are the final stable decay products? From the evidence,(b) is the $A_{2}^{+}$ particle, a fermion or a boson and (c) is it a meson or a baryon?(d) What is its baryon number?

Which conservation law is violated in each of these proposed decays? Assume that the initial particle is stationary and the decay products have zero orbital angular momentum.

(a) $μ^{-} \to e^{-} + ν_{μ}$ ; (b) $μ^{-} \to e^{+} + ν_{e} + {v^{-}}_{μ}$ ;(c) $μ^{+} \to π^{+} + ν_{μ^{+}}$

The wavelength at which a thermal radiator at temperature $T$ radiates electromagnetic waves most intensely is given by Wien’slaw:localid="1663130298382" $λ_{m a x} = (2898 μ m . K) / T$ (a) Show that the energy $E$ of a photon corresponding to that wavelength can be computed from

$E = (4 .28 \times 10^{- 10} \frac{MeV}{K}) T$

(b) At what minimum temperature can this photon create an electron-positron pair(as discussed in Module 21-3)?

See all solutions

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