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

  • Thread starter Thread starter patric44
  • Start date Start date
  • Tags Tags
    Shift States
patric44
Messages
308
Reaction score
40
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.
 
Physics news on Phys.org
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/
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top