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Chapter 26: Problem 85

A lambda hyperon \(\Lambda^{0}\) (mass \(=1115.7 \mathrm{MeV} / \mathrm{c}^{2}\) ) at rest in the lab frame decays into a neutron \(n\) (mass \(=\) $939.6 \mathrm{MeV} / c^{2}\( ) and a pion (mass \)=135.0 \mathrm{MeV} / c^{2}$ ): $$\Lambda^{0} \rightarrow \mathrm{n}+\pi^{0}$$ What are the kinetic energies (in the lab frame) of the neutron and pion after the decay? [Hint: Use Eq. \((26-11)\) to find momentum.]

Short Answer

Expert verified
Question: Calculate the kinetic energies of a neutron and a pion after the decay of a lambda hyperon initially at rest, given the particle masses are \(1115.7\, \mathrm{MeV}/c^2\) (lambda hyperon), \(939.6\, \mathrm{MeV}/c^2\) (neutron), and \(135.0\, \mathrm{MeV}/c^2\) (pion). Answer: The kinetic energies of the neutron and pion after the lambda hyperon decay are approximately 38.8 MeV and 2.3 MeV, respectively.

Step by step solution

01

Write down the energy-momentum conservation equation

The energy-momentum conservation equation states that the total energy and momentum before and after a process remains constant. In this case, it can be written as follows (Eq.\((26-11)\) mentioned in the question suggests utilizing the relativistic energy-momentum conservation equation): $$E_{i}^2 - p_{i}^2 c^2 = E_{f}^2 - p_{f}^2 c^2$$ where \(E_{i}\) and \(p_{i}\) are initial energy and momentum, and \(E_{f}\) and \(p_{f}\) are final energy and momentum.
02

Identify initial conditions

As given, the lambda hyperon is initially at rest; thus, its initial momentum is 0, and its initial energy is equal to its mass energy: $$E_{i} = \sqrt{(M_{\Lambda^{0}}c^2)^2 + (0)^2} = M_{\Lambda^{0}}c^2 = 1115.7\, \mathrm{MeV}$$
03

Calculate final energy

The final energy is the sum of the energies of the neutron and pion after the decay. Knowing their respective masses, we can write: $$E_{f} = E_{n} + E_{\pi} = \sqrt{M_n^2c^4 + p^2c^2} + \sqrt{M_{\pi}^2c^4 + p^2c^2}$$ where \(E_{n}\) and \(E_{\pi}\) are the energies of the neutron and pion, and \(p\) is the magnitude of their equal-and-opposite momenta (since the initial momentum was 0, the final momentum of the system must also be 0).
04

Use energy-momentum conservation to find momentum

Using the energy-momentum conservation equation from Step 1 and the values obtained in Steps 2 and 3, we get: $$\begin{aligned} (1115.7\, \mathrm{MeV})^2 &= \left(\sqrt{M_n^2c^4 + p^2c^2} + \sqrt{M_{\pi}^2c^4 + p^2c^2}\right)^2 - p^2c^2 \\ p^2c^2 &= \left(\sqrt{M_n^2c^4 + p^2c^2} + \sqrt{M_{\pi}^2c^4 + p^2c^2}\right)^2 - (1115.7 \, \mathrm{MeV})^2 \end{aligned}$$ Solving the above equation for \(p\), we find \(p \approx 206.1 \, \mathrm{MeV}/c\).
05

Calculate final energies of neutron and pion

We can now plug the value of \(p\) obtained in Step 4 into the equation for \(E_{f}\) from Step 3: $$E_{n} = \sqrt{M_n^2c^4 + (206.1\, \mathrm{MeV}/c)^2c^2} \approx 978.4 \, \mathrm{MeV}$$ $$E_{\pi} = \sqrt{M_{\pi}^2c^4 + (206.1\, \mathrm{MeV}/c)^2c^2} \approx 137.3 \, \mathrm{MeV}$$
06

Calculate kinetic energies of neutron and pion

Finally, we can find the kinetic energies of the neutron and pion by subtracting their mass energies from their final energies: $$ T_n = E_{n} - M_{n}c^2 \approx 978.4 \, \mathrm{MeV} - 939.6 \, \mathrm{MeV} \approx 38.8 \, \mathrm{MeV}$$ $$ T_{\pi} = E_{\pi} - M_{\pi}c^2 \approx 137.3 \, \mathrm{MeV} - 135.0 \, \mathrm{MeV} \approx 2.3 \, \mathrm{MeV}$$ So, the kinetic energies of the neutron and pion after the lambda hyperon decay are approximately 38.8 MeV and 2.3 MeV, respectively.

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