Physics
Physics
5th Edition
ISBN: 9781260487008
Author: GIAMBATTISTA, Alan
Publisher: MCGRAW-HILL HIGHER 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 energies (in the lab frame) of the neutron and pion after decay is 5.7MeV and kinetic energy pion after decay is 35.4MeV

Explanation of Solution

Write the equation for the relativistic energy of a particle.

  (pc)2=K2+2KE0        (I)

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        (II)

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

Write the expression for the relativistic energy of a neutron.

  (pπc)2=Kπ2+2KπE0π        (III)

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

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

Equate equation (II) and (III)

  Kn2+2KnE0n=Kπ2+2KπE0π        (IV)

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

mΛc2=mnc2+mπc2+Kn+Kπ        (V)

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

Rearrange above equation to get Kn+Kπ

  (mΛmnmπ)c2=Kn+Kπ=K        (VI)

Here, K  is the total kinetic energy.

Rearrange the equation K=Kn+Kπ to get Kn

  Kn=KKπ        (VII)

Substitute (mΛmnmπ)c2 for K in expression (VII)                            Kn=(mΛmnmπ)c2Kπ        (VIII)

Substitute expression (VII) in expression (IV) and solve for Kπ.

    (KKπ)2+2(KKπ)E0n=Kπ2+2KπE0πK2+2KKπ+Kπ2+2KE0n=Kπ2+2KπE0π2Kπ(K+E0n+E0π)=K(K+2E0n)Kπ=K(K+2E0n)2(K+E0n+E0π)        (IX)

Substitute (mΛmnmπ)c2 for K, mnc2 for E0n, and mπc2 for E0π in expression (IX),

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

Conclusion:

Substitute the 1115.7MeV/c2 for mΛ939.6MeV/c2 for mn, 135.0MeV/c2 for mπ in expression (x)

  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)c2+(939.6MeV/c2)c2+(135.0MeV/c2)c2)=35.4MeV

Substitute the 1115.7MeV/c2 for mΛ939.6MeV/c2 for mn, 135.0MeV/c2 for mπ , and 35.4MeV for Kπ in expression (VII)

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

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

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