Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 26, Problem 92P
To determine

The kinetic energies (in the lab frame) of the neutron and pion after decay.

Expert Solution & Answer
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Answer to Problem 92P

The kinetic energy (in the lab frame) of the neutron after decay is 5.7MeV and kinetic energy pion after decay is 35.4MeV.

Explanation of Solution

Write the expression for the relativistic energy of a particle.

  (pc)2=K2+2KE0

Here, p is the momentum of the particle, K is the kinetic energy of the particle and E0 is the rest energy of the particle.

Write the expression for the relativistic energy of a neutron.

  (pnc)2=Kn2+2KnE0n                                                                                         (I)

Here, pn is the momentum of neutron, Kn is the kinetic energy of neutron and E0n is the rest energy of neutron.

Write the expression for the relativistic energy of a pion.

  (pπc)2=Kπ2+2KπE0π                                                                                      (II)

Here, pπ is the momentum of pion, Kπ is the kinetic energy of pion and E0π is the rest energy of pion.

Since lambda hyperon is at rest in the lab frame, by conservation of momentum pn=pπ.

Equate equation (I) and (II).

  Kn2+2KnE0n=Kπ2+2KπE0π                                                                            (III)

Thus, total energy of the system is conserved. Apply conservation of energy for the decay process.

  mΛc2=mnc2+mπc2+Kn+Kπ

Here, mΛ is the mass of lambda hyperon, mn is the mass of neutron, mπ is the mass of pion and c is the speed of light in vacuum.

Rearrange above equation to get Kn+Kπ.

  (mΛmnmπ)c2=Kn+Kπ=K                                                                   (IV)

Here, K is the total kinetic energy.

Rearrange the equation K=Kn+Kπ to get Kn.

  Kn=KKπ

Substitute (mΛmnmπ)c2 for K in above equation to get Kn.

  Kn=(mΛmnmπ)c2Kπ                                                                             (V)

Substitute KKπ for Kn in equation (III) to get Kπ.

  (KKπ)2+2(KKπ)E0n=Kπ2+2KπE0πK22KKπ+Kπ2+2KE0n2KπE0n=Kπ2+2KπE0π2KKπ+2KπE0n+2KπE0π=2KE0n+K22Kπ(K+E0n+E0π)=K(K+2E0n)

Solve above equation to get Kπ.

  Kπ=K(K+2E0n)2(K+E0n+E0π)

Substitute (mΛmnmπ)c2 for K , mnc2 for E0n and mπc2 for E0π in above equation to get expression of Kπ in terms of masses of particles.

  Kπ=(mΛmnmπ)c2((mΛmnmπ)c2+2mnc2)2((mΛmnmπ)c2+mnc2+mπc2)                                     (VI)

Conclusion:

Substitute 1115.7MeV/c2 for mΛ , 939.6MeV/c2 for mn, 135.0MeV/c2 for mπ in above equation to get Kπ.

Kπ=(1115.7MeV/c2939.6MeV/c2135.0MeV/c2)c2((1115.7MeV/c2939.6MeV/c2135.0MeV/c2)c2+2×939.6MeV/c2×c2)2[(1115.7MeV/c2939.6MeV/c2135.0MeV/c2)+939.6MeV/c2×c2+135.0MeV/c2×c2]Kπ=(1115.7MeV939.6MeV135.0MeV)((1115.7MeV939.6MeV135.0MeV)+2×939.6MeV)2((1115.7MeV939.6MeV135.0MeV)+939.6MeV+135.0MeV)=35.4MeV

Substitute 1115.7MeV/c2 for mΛ , 939.6MeV/c2 for mn, 135.0MeV/c2 for mπ and 35.4MeV for Kπ in equation (V) to get Kn.

  Kn=(1115.7MeV/c2939.6MeV/c2135.0MeV/c2)c235.4MeV=5.7MeV

Therefore, the kinetic energy (in the lab frame) of the neutron after decay is 5.7MeV and kinetic energy pion after decay is 35.4MeV.

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