Calculate the isotopic shift of the analogue states for Cr and Mn

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SUMMARY

The discussion focuses on calculating the isotopic shift of isobaric analogue states for Chromium (Cr) and Manganese (Mn) using specific equations derived from nuclear physics principles. The key formulas provided include the energy shift equations, which are essential for determining the isotopic shift in the multiplet (B-C-N). The user seeks clarification on the availability of necessary data, specifically the values of Ex> and Ex< for Cr and Mn, and confirms the terminology distinction between isobaric and isotopic analogue states. Additionally, the discussion highlights the stable isotopes of Cr and Mn, emphasizing their relevance in nuclear decay processes.

PREREQUISITES
  • Understanding of nuclear physics concepts, particularly isospin multiplets.
  • Familiarity with isotopic and isobaric terminology in nuclear reactions.
  • Knowledge of energy calculations in nuclear physics, including binding energy.
  • Basic understanding of radionuclides and their decay processes.
NEXT STEPS
  • Research the concept of isobaric analogue states in nuclear physics.
  • Study the equations for calculating isotopic shifts in nuclear reactions.
  • Explore advanced nuclear physics textbooks that cover isospin and multiplet theory.
  • Investigate the decay chains of radionuclides, focusing on 47Cr and 47Mn.
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Nuclear physicists, students studying nuclear reactions, and researchers interested in isotopic shifts and isobaric analogue states in Cr and Mn.

patric44
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Homework Statement
given the isotopic doublet Cr-Mn ,calculate the isotopic shift of the analogue states given the information below.
Relevant Equations
Ex> = Ex<+Exe+delta , where delta is the shift
hi guys
my nuclear physics professor gave us a hand written notes about a the isospin multiplets of elements, the notes provides a brief not clear introduction to the topic with some formulas for calculations, as following
$$
E_{xe} = \Delta\;E_{B}+\Delta\;E_{c}
$$
$$
\Delta\;E_{B}= E_{B}(z>,A)-E_{B}(z<,A)
$$
$$
\Delta\;E_{c}= 1.45*\frac{z>}{A^{1/3}}-1.03
$$
and finally
$$
E_{x>} = E_{x>}+E_{xe}+\delta
$$
the notes are very old and written by hand so its not clear, but i systematically manged to get through how to calculate the isotopic shift for the multiplet (B-C-N) in the example provided in the notes, However i found the following problem asking the same question :
1657215104216.png

is there is something missing in the given data, like where is Ex> and Ex< for Cr or Mn to calculate the isotopic shift??
is there are anyone familiar with this topic in nuclear physics , i will appreciate any help, also i was not able to find this topic as its with the given equations in any nuclear physics book that i knew, most of them talk about neutron proton isospin and T,Tz.
 
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I just want to make sure, is it called the isobaric analogue states or the isotopic analogue states?, and can anyone suggest to me a book or some notes on the subject, i would really appreciate it
 
Definition - isobar, in nuclear physics, any member of a group of atomic or nuclear species all of which have the same mass number — that is, the same total number of protons and neutrons. Isotope refers to nuclides of the same element, i.e., some number of protons (and electrons in a neutral atom) and differing number of neutrons. Each nuclide 47Cr and 47Mn are radionuclides in general, and are isotopes (radioisotopes) of the respective elements, but they are isobars to one another. Both decay by electron capture. 47Mn (88 ms) -> 47Cr (500 ms) -> 47V (32.6 min) -> 47Ti (stable).

Cr has three stable isotopes/nuclides, 52Cr, 53Cr and 54Cr, with 50Cr is more or less stable due to a very long half life, est. >1.3E+18 y.

Mn has one stable isotope/nuclide 55Mn.

Confirm with https://www.nndc.bnl.gov/nudat3/
 
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