Evaluate the energies of muonic K X rays in Fe

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SUMMARY

The discussion focuses on evaluating the energies of muonic K X rays in iron (Fe) using a one-electron model and a point nucleus assumption. The key equation for energy difference, \(\Delta E\), is derived from the relationship between the 1s state energy of a point nucleus and a uniformly charged nucleus. The muon, being 207 times the mass of an electron, results in a Bohr radius of \(a_{\mu}=\frac{1}{207}a_{o}\). The final energy calculation, incorporating Mosley's law and the transition from the P(3/2) to the 1s(1/2) state, yields expected energy values around 1.3 MeV with a margin of error of ±0.1 MeV.

PREREQUISITES
  • Understanding of muonic atoms and their properties
  • Familiarity with Bohr model calculations
  • Knowledge of Mosley's law for X-ray energies
  • Basic concepts of nuclear physics and isotopes
NEXT STEPS
  • Study the derivation and applications of Mosley's law in X-ray spectroscopy
  • Explore the differences between muonic and electronic orbits in detail
  • Research the properties of muons and their interactions in atomic physics
  • Investigate the energy levels of various isotopes of iron (Fe) and their implications in nuclear physics
USEFUL FOR

This discussion is beneficial for physicists, particularly those specializing in atomic and nuclear physics, as well as students working on advanced topics in quantum mechanics and particle physics.

Demon117
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Homework Statement


Using a 1 electron model, evaluate the energies of the muonic K X rays in Fe assuming a point nucleus.


Homework Equations


The \Delta E is the difference between energy of the 1s state in an atom with a "point" nucleus and the 1s energy in an atom with uniformly charged nucleus. The equation is found to be

\Delta E= \frac{2Z^{4}e^{2}R^{2}}{5(4 \pi \varepsilon_{o})a_{o}^{3}}

R=R_{o}A^{1/3}


The Attempt at a Solution


The muon is a particle identical to the electron in all characteristics except its mass, which is 207 times the electron mass. Since the Bohr radius depends only on the mass, the muonic orbits have 1/207 the radius of the corresponding electron orbits.

So, this means that the Bohr radius of a muon will be a_{\mu}=\frac{1}{207}a_{o}. By that logic, we should have:


\Delta E= \frac{2Z^{4}e^{2}(R_{o}A^{1/3})^{2}}{5(4 \pi \varepsilon_{o})(\frac{1}{207}a_{o})^{3}}

By substitution. . . . when I do this, I end up with energies far below what is expected in the muonic k x rays in some Fe isotopes, maybe my thinking is incorrect. Could someone lend a hand? I have been doing a lot of HW this weekend and I think I am at my breaking point.
 
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matumich26 said:

Homework Statement


Using a 1 electron model, evaluate the energies of the muonic K X rays in Fe assuming a point nucleus.


Homework Equations


The \Delta E is the difference between energy of the 1s state in an atom with a "point" nucleus and the 1s energy in an atom with uniformly charged nucleus. The equation is found to be

\Delta E= \frac{2Z^{4}e^{2}R^{2}}{5(4 \pi \varepsilon_{o})a_{o}^{3}}

R=R_{o}A^{1/3}


The Attempt at a Solution


The muon is a particle identical to the electron in all characteristics except its mass, which is 207 times the electron mass. Since the Bohr radius depends only on the mass, the muonic orbits have 1/207 the radius of the corresponding electron orbits.

So, this means that the Bohr radius of a muon will be a_{\mu}=\frac{1}{207}a_{o}. By that logic, we should have:


\Delta E= \frac{2Z^{4}e^{2}(R_{o}A^{1/3})^{2}}{5(4 \pi \varepsilon_{o})(\frac{1}{207}a_{o})^{3}}

By substitution. . . . when I do this, I end up with energies far below what is expected in the muonic k x rays in some Fe isotopes, maybe my thinking is incorrect. Could someone lend a hand? I have been doing a lot of HW this weekend and I think I am at my breaking point.

Realized I need to use Mosley's law for this and note that the transition is from the P(3/2) to the 1s(1/2) state, and adjusting for the muon mass in the Rydberg constant the energy falls right within where we expect, about 1.3 MeV +/- .1 MeV.
 

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