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Chapter 30: Problem 27

A sigma baryon at rest decays into a lambda baryon and a photon: $\Sigma^{0} \rightarrow \Lambda^{0}+\gamma .$ The rest energies of the baryons are given by \(\Sigma^{0}=1192 \mathrm{MeV}\) and \(\Lambda^{0}=1116 \mathrm{MeV} .\) What is the photon wavelength? [Hint: Use relativistic formulas and be sure momentum is conserved as well as energy.]

Short Answer

Expert verified
Question: Determine the wavelength of a photon resulting from the decay of a sigma baryon at rest into a lambda baryon and a photon, given the rest energies of the sigma baryon \(E_{\Sigma^{0}} = 1192 \,\mathrm{MeV}\) and the lambda baryon \(E_{\Lambda^{0}} = 1116 \,\mathrm{MeV}\). Answer: The wavelength of the photon resulting from the decay is approximately \(\lambda \approx 6.651\times 10^{-15}\,\mathrm{m}\).

Step by step solution

01

Define the energy conservation equation.

During the decay process, energy is conserved. Therefore, the rest energy of the sigma baryon turns into the sum of the energies of the lambda baryon and the photon. In equation form, this relation can be expressed as: \(E_{\Sigma^{0}}=E_{\Lambda^{0}}+E_{\gamma}\)
02

Define the momentum conservation equation.

During the decay process, momentum is conserved. Since the sigma baryon is at rest, its initial momentum is zero. After the decay, the final momenta of the lambda baryon and the photon should sum up to zero (they should have equal magnitudes and opposite directions). In equation form, this relation can be expressed as: \(p_{\Lambda^{0}}=p_{\gamma}\)
03

Express the energy and momentum of the photon in terms of its wavelength and lambda baryon energy.

We can use the Planck energy equation to relate the energy of the photon to its frequency (\(E_{\gamma}=h\nu\)) and its wavelength (\(\lambda=c/\nu\)): \(E_{\gamma}=\frac{hc}{\lambda}\) The momentum of the photon is related to its wavelength by the de Broglie equation: \(p_{\gamma}=\frac{h}{\lambda}\)
04

Solve the energy conservation equation for the energy of the lambda baryon.

Rearranging the energy conservation equation (step 1), we can write the energy of the lambda baryon in terms of the rest energies of the baryons and the photon wavelength: \(E_{\Lambda^{0}}=E_{\Sigma^{0}}-\frac{hc}{\lambda}\)
05

Solve the momentum conservation equation for the momentum of the lambda baryon.

We can write the following relations for the lambda baryon energy and momentum using the relativistic formulas: \(E_{\Lambda^{0}}=\frac{m_{\Lambda^{0}}c^2}{\sqrt{1-v_{\Lambda^{0}\,2}/c^2}}\) \(p_{\Lambda^{0}}=\frac{m_{\Lambda^{0}}v_{\Lambda^{0}}}{\sqrt{1-v_{\Lambda^{0}\,2}/c^2}}\) From the momentum conservation equation (step 2), the momentum of the lambda baryon and the photon are equal: \(p_{\Lambda^{0}}=\frac{h}{\lambda}\)
06

Solve for the lambda baryon velocity and energy.

Using the equations from steps 5, we can solve for the lambda baryon velocity (\(v_{\Lambda^{0}}\)) and its energy (\(E_{\Lambda^{0}}\)).
07

Solve for the photon wavelength.

Substitute the lambda baryon energy from step 6 into the adjusted energy conservation equation from step 4: \(E_{\Sigma^{0}}-E_{\Lambda^{0}}=\frac{hc}{\lambda}\) Now we can solve for the wavelength of the photon, \(\lambda\): \(\lambda=\frac{hc}{E_{\Sigma^{0}}-E_{\Lambda^{0}}}\)
08

Plug in the given values and calculate the photon wavelength.

We are given the rest energies of the sigma and lambda baryons in MeV, and we can convert them into joules by multiplying by \(1.60218\times 10^{-13}\mathrm{J/MeV}\). Using the speed of light (\(c=3\times 10^8 \mathrm{m/s}\)) and the Planck constant (\(h=6.626\times 10^{-34} \mathrm{Js}\)), we can calculate the photon wavelength: \(\lambda = \frac{(6.626\times 10^{-34}\,\mathrm{Js})(3\times 10^8\,\mathrm{m/s})}{[(1192\,\mathrm{MeV})(1.60218\times10^{-13}\,\mathrm{J/MeV})-(1116\,\mathrm{MeV})(1.60218\times10^{-13}\,\mathrm{J/MeV})]}\) Completing this calculation, we find the photon wavelength to be approximately: \(\lambda\approx 6.651\times 10^{-15}\,\mathrm{m}\)

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