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Chapter 26: Problem 42

A lambda hyperon \(\Lambda^{0}\) (mass \(=1115 \mathrm{MeV} / c^{2}\) ) at rest decays into a neutron \(\mathrm{n}\) (mass \(=940 \mathrm{MeV} / \mathrm{c}^{2}\) ) and a pion (mass \(=135 \mathrm{MeV} / c^{2}\)): $$\Lambda^{0} \rightarrow \mathrm{n}+\pi^{0}$$ What is the total kinetic energy of the neutron and pion?

Short Answer

Expert verified
Answer: The total kinetic energy of the neutron and the pion after the decay is 175 MeV.

Step by step solution

01

Write down the energy-momentum conservation equations

In this particle decay, energy and momentum must be conserved. For the energy conservation, we have: $$E_{\Lambda^0}= E_n + E_{\pi^0}$$ For the momentum conservation we have: $$\vec{p}_{\Lambda^0} = \vec{p_n} + \vec{p_{\pi^0}}$$ Since the lambda hyperon is initially at rest, its momentum is zero. Thus, the magnitudes of momenta of the neutron and pion must be equal and opposite: $$|\vec{p_n}| = |\vec{p_{\pi^0}}|$$
02

Use the energy-mass-momentum relation

To relate the energy of a particle with its mass and momentum, we use the following equation: $$E = \sqrt{(pc)^2 + (mc^2)^2}$$ For the neutron and the pion, we have: $$E_n = \sqrt{(p_nc)^2 + (m_nc^2)^2}$$ $$E_{\pi^0} = \sqrt{(p_{\pi^0}c)^2 + (m_{\pi^0}c^2)^2}$$
03

Substitute energy-mass-momentum relations into energy conservation equation

Now we substitute the relations we obtained in step 2 into the energy conservation equation from step 1: $$E_{\Lambda^0} =\sqrt{(p_nc)^2 + (m_nc^2)^2}+ \sqrt{(p_{\pi^0}c)^2 + (m_{\pi^0}c^2)^2}$$
04

Eliminate momentum terms from the equation

We can eliminate the momentum terms from the equation because the magnitudes of the momenta of the neutron and the pion are equal and opposite (from step 1): $$E_{\Lambda^0} =\sqrt{(p_n c)^2 + (m_n c^2)^2}+ \sqrt{(p_{n}c)^2 + (m_{\pi^0}c^2)^2}$$
05

Calculate the energy difference

The kinetic energy is the difference between the total energy and the rest mass energy of a particle. Thus, the total kinetic energy of the neutron and the pion is: $$KE_{total} = E_n + E_{\pi^0} - m_nc^2 - m_{\pi^0}c^2$$ We can now substitute the energies and masses given in the problem: $$KE_{total} = (E_{\Lambda^0} - m_{\Lambda^0}c^2) - m_nc^2 - m_{\pi^0}c^2$$ $$KE_{total} = (1115 - 1115) \mathrm{MeV} - 940 \mathrm{MeV} - 135 \mathrm{MeV}$$
06

Compute total kinetic energy

Now we can compute the total kinetic energy of the neutron and the pion after the decay: $$KE_{total} = - 960 \mathrm{MeV} = 175 \mathrm{MeV}$$ The total kinetic energy of the neutron and the pion after the decay is 175 MeV.

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Most popular questions from this chapter

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