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Unit conversion J/K into m/s

  1. Dec 13, 2013 #1

    Rajini

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    Dear All,

    J/K into m/s ?
    Or
    kgJs/m into kg?

    thanks for your help!
    Rajini.
     
    Last edited: Dec 13, 2013
    Rajini, Dec 13, 2013
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  3. Dec 13, 2013 #2

    Cthugha

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    There is no conversion because these are completely different units.

    J is kg m^2/s^2, so things do not match.

    J/K is the typical unit for heat capacity or entropy, while m/s gives a velocity.

    Is there something wrong with the calculations or are non-SI units used? In natural units both the Boltzmann constant and c are equal to a dimensionless 1, while they have units of J/K and m/s in SI-units, respectively.
     
    Cthugha, Dec 13, 2013
  4. Dec 13, 2013 #3

    dauto

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    These two units are incommensurable. They cannot be converted.
     
    dauto, Dec 13, 2013
  5. Dec 15, 2013 #4

    Rajini

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    seems one can do

    Hello,
    Sorry I did not give more information!
    Please see this equation (taken from J. Solid state Commun, Vol. 33, p.361, 1980):
    dIS/dT=-3EγkB/(2Mc2).
    Here dIS/dT is in mm/(sK). This value can be obtained from plot.
    Eγ=14412.497 eV.
    M=56.9353987/(1000×6.02214179×1023) kg.
    kB=8.6173324 eV/k=1.3806488×10-23 J/K.
    c=299792458 m/s.
    I measure IS vaules (in mm/s) for various temperatures (in K) and using the above formula I can plot IS values on y-axis and T on x-axis. Now find slope, i.e. dIS/dT (unit will be mm/(sK)).
    Now can determine the M vaues in kg or amu.
    But on considering the unit I am getting confused!!
    After rearranging the formula to
    M=-3EγkB/(2c2(dIS/dT)).
    After substituing the units I get
    eV eV s2 s K/(K m2 mm)=eV2s3/(m2mm)=eV2s3 1000/m3.
    Now I get M=2.0728216×10-14eV2s3/m3.
    How to balance the units, how to convert the units in right hand side to get unit in kg??
    Please help.

    Thanks and regards, Rajini.
    NOTE:
    Useful conversion: 1 mm/s=48.075×10-9 eV.
     
    Last edited: Dec 15, 2013
    Rajini, Dec 15, 2013
  6. Dec 16, 2013 #5

    Cthugha

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    This reference does not exist. There is no J. Solid state Commun., so I suppose you mean Solid State communications which had a volume 33 in 1980, but there is no article starting on p. 361. The article ending on p. 361, "PHASE TRANSITION IN THE TRANSVERSE FIELD DEPENDENT ON THE DEFORMATION" does not contain your equation. Could you please provide the name of the article, so we can find it.

    This also does not work as the units do not add up, so I assume that there is something set to 1 which makes it work. IF your above conversion works, also your equation works out.
     
    Cthugha, Dec 16, 2013
  7. Dec 16, 2013 #6

    Rajini

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    Sorry I wrongly informed!!!
    The correct one..see eq. 2 and table II..where the author computed M
    Journal of Solid State Chemistry
    Volume 33, Issue 3, 15 July 1980, Pages 361–369.
    I already verified eq. 1 and 3..and they are correct in balancing the units..
    PS:Finally I got 8.9686×10-6/(dIS/dT)s/m.
    After this I dont how to proceed!!!
     
    Last edited: Dec 16, 2013
    Rajini, Dec 16, 2013
  8. Dec 16, 2013 #7

    Cthugha

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    Oh, this is Mößbauer stuff.

    If I remember correctly (I had to do some standard experiment while I was a student, so my memory might fail me here - please double check what I say here) they perform energy variations by applying Doppler shifts:
    [tex]E_\gamma (v)=E_0 (1+\frac{v}{c})[/tex]
    and it is quite common for some groups to only give the velocity when they actually mean the energy. So you might want to check, whether the values work out when you convert the velocities you get to energies that way.
     
    Cthugha, Dec 16, 2013
  9. Dec 16, 2013 #8

    Rajini

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    Yes, one and only noble prize winner till now I saw in Munchen in 2009.


    [tex]E_\gamma (v)=E_0 (1+\frac{v}{c})[/tex] Yes I used this formula for mm/s to eV conversion.
    E.g., Doppler shifted energy ED=vEγ/c. Here by taking v=1mm/s one can get 4.8075×10-8, i.e. 1mm/s=4.8075×10-8eV. Thinking of directly asking the author o:).

    Thanks, Rajini.
     
    Last edited: Dec 16, 2013
    Rajini, Dec 16, 2013
  10. Dec 20, 2013 #9

    Rajini

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    Hello all,
    I solved!!
    From my previous post we have
    M=-3EγkB/2c2[dIS/dT].
    Now taking Eγ in eV
    kB in eV/K
    c in m/s
    dIS/dT in mm/(sK)
    M=eV eV s2 s K/(K m2 mm)
    = eV eV s2 s K/(K m2 10-3m)
    Include 'kg' in numerator and denominator so J=kg m2/s2, which will cancel with one eV
    M=eV s kg/m
    substitute m/s in eV
    Finally
    M=4.16×10-5/(dIS/dT) amu. (dIS/dT values in m/(sK)).
    I also checked with that paper and got the correct exact values reported there.
    regards, raja.
     
    Rajini, Dec 20, 2013
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