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Rest energy of 238Uranium and gamma factor

  1. Aug 5, 2015 #1
    Hi everyone!

    So, I know that for a certain Synchrotron I have an output value of 1.5GeV/u for Uranium-238 particles, with a charged state of 28+ (## ^{238}U^{+28} ##). For this ion I would like to calculate the rest energy and the associated relativistic gamma factor. My approach was the following:

    • Rest energy (ignoring the mass of electrons):
    For the rest energy, I considered that Uranium-238 is made of 238 neutrons and protons in total, and I just multiplied this value by the atomic mass number:

    ## E_0 = mc^2= (238 \cdot amu)c^2 = (238 \cdot 1.661\times 10^{-27})c^2 = 3.56 \times 10^{-8} J##

    For eV units we just divide by the electron charge and get:

    ## E_0 = 3.56 \times 10^{-8} / e = 2.221 \times 10^{11} eV = 222.1 GeV##
    • Total energy:
    Now, for the total energy of the particle, I considered that since it has 1.5GeV/u, meaning 1.5GeV per nucleon, I just multiply it by 238 and I get the total energy:

    ## E_{total} = 1.5\times 238 = 357 GeV ##
    • Gamma factor:
    Finally, for the gamma factor, this is just the ratio between total energy and rest energy:

    ##\gamma = E_{total}/E_0 = 357/222.1 = 1.607 ##

    Do you think this is the right approach to do it?
    What I'm not sure here is if the 1.5GeV/u is defining the total energy, or just the kinetic energy... Which would have an impact on the gamma factor.

    What is more common to state in particle accelerators? Kinetic or total energy?

    Thanks in advance! :)
  2. jcsd
  3. Aug 5, 2015 #2


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    Is this homework, or is it a real-world problem? If it's a real-world problem, can't you ask whoever you're working with whether they mean the total energy or just kinetic energy? If it's homework, please read this: https://www.physicsforums.com/help/homeworkhelp/

    Your method of finding the rest mass is probably good to about 2 sig figs, but why not just look it up rather than trying to estimate it?
  4. Aug 5, 2015 #3
    Hi bcrowell,

    This is a real-world problem. But, unfortunately I have no access to the person who wrote this.
    In reality, what I'm trying to figure out is the relation between these two numbers presented on a table for emittance in this synchrotron for the Uranium-238 ion:

    εx (norm.) [mm mrad] = 32
    εx (phys.) [mm mrad] = 47

    I thought that the number εx (phys.) would be the emittance, and εx (norm.) the normalized emittance, which are related by:

    ## \epsilon_{norm} = \gamma \beta \epsilon ##

    So, I was trying to see if this is what they meant. But, my calculations for ## \gamma \beta ## for this ion don't match their ratio, which is around ## \gamma \beta = 32/47 = 0.681 ##.

    I have ## \gamma = 1.607 ## and ## \beta = 0.783 ## which gives ## \gamma \beta = 1.258 ##.

  5. Aug 5, 2015 #4


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    Yes, your calculation of gamma looks right to me, if the given energy is total energy. If you interpret it as kinetic energy, the disagreement with the expected value of ##\gamma\beta## becomes worse, right?

    Surely if you're working with a synchrotron you have access to accurate information about its energy...?
  6. Aug 5, 2015 #5
    Yes, the disagreement is even worse.
    The synchrotron is not built yet, these are some design parameters. These values came from some tables, but no explanation on them :frown:

  7. Aug 5, 2015 #6


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    Other specifications like the dipole magnet strength and the size could help. Data about the preaccelerator could be interesting as well.
  8. Aug 5, 2015 #7
    Hi mfb,

    That I know, the Synchrotron has a magnetic rigidity of 100 Tesla meter and the Dipole magnet field is around 2 Tesla with a curvature of around 50meter. But, how can I go from here to the values of emittance? These emittance values are on the output of the synchrotron.

    Well, I just wanted to find out what exaclty do they mean by these different definitions of emittance they present on some tables: εx (norm.) [mm mrad] and εx (phys.) [mm mrad]

  9. Aug 5, 2015 #8


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    You won't get the emittance puzzle solved, but it can help with the beam energy.
    92 e * 100 Tm/(2pi) = 439 GeV/c, or 1.84 GeV/nucleon (total momentum).
    Therefore, ##\gamma \beta \approx \frac{1.84}{0.938} \approx 1.97##.
  10. Aug 5, 2015 #9
    Hi again mfb,

    Thanks for the reply! I didn't quite get what you did there... from ## B \rho = \frac{p}{e} ## how did you get that? I'm puzzled about the 2pi factor...

    thanks! :)
  11. Aug 5, 2015 #10


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    If the 100 Tesla meter refer to the radius, the values are way off. For the circumference it would make sense.
    The accelerator could run with a weaker field for your uranium, of course, then the design values don't help.
  12. Aug 5, 2015 #11
    Yes, it's for the circumference.
    Actually they state that for heavy ions they will use a ## B \rho = 90 ## and a dipole field of 1.9T. The bending radius of the dipole is 47m.

    For Synchrotrons it's more common to define the emittance or the normalized emittance?

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