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Linewidth and time

  1. Jun 4, 2010 #1
    Hello all,

    is there any possibility to express line width [tex]\Gamma[/tex] of the nuclear level in second (s) or inverse second ?
    For eg. for 57Fe,
    [tex]\Gamma=4.6413[/tex] neV = 0.09654 mm/s for the 1st excited level of 57Fe.
    I want to how one can convert to second or inverse second.
    PS: some conversion factors that i know are: 1 mm/s = 48.075 neV and 1 eV = 8065.5 cm-1
    thanks
     
  2. jcsd
  3. Jun 4, 2010 #2
    Use Planck's constant

    h/2π = 6.582118 x 10-16 eV-seconds

    Bob S
     
  4. Jun 4, 2010 #3

    Meir Achuz

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    I use the two factors: 1=197.32 MeV-fm
    and 1=2.9979...X10^23 fm/sec.

    They give 1 MeV=197.32/( 2.9979 X 10^23) /sec.
    divide by 10^15 to get it in neV.
     
    Last edited: Jun 4, 2010
  5. Jun 4, 2010 #4
    hello,
    i think i have not supplied some information.
    In a book i found the following relation:
    [tex]\Gamma/\omega_N<1[/tex]. Here [tex]\Gamma[/tex] is nuclear line width and [tex]\omega_N[/tex] is Larmor precession frequency.
    But i prefer to write as [tex]\tau_N<(1/\Gamma)[/tex]---is this correct?[[tex]\tau_N=(\omega_N)^{-1}[/tex]]
    What i understand is the unit of [tex]\tau_N[/tex] (obviously unit is s) and [tex]\Gamma[/tex] should be same. Also [tex]\Gamma[/tex] can take unit as mm/s, eV, etc and [tex]\omega_N[/tex] may be in Hz i guess. But we also know that inverse of [tex]\omega_N[/tex] has a unit in second.
    But the gamma value is fixed Mössbauer related constant. For 57Fe and usually given in eV or neV or mm/s. and mean lifetime is 141 ns.
    thanks again
     
  6. Jun 4, 2010 #5
    Hi-
    In your original post, you state that the energy uncertainty is 4.6413 x 10-9 ev, and in the above post that the mean lifetime is 141 x 10-9 seconds. The product is 6.544 x 10-16 eV-seconds, very close to the value for the value of Plancks constant (h/2π) = 6.582118 x 10-16 eV-seconds, given in post #1. This is a hint.

    Bob S

    [Added] See http://en.wikipedia.org/wiki/Pound–Rebka_experiment

    δE/E = 4.64 x 10-9 eV / 14 KeV = 3.3 x 10-13

    δv/c = 0.09564 mm per sec / 3 x 1011 mm per sec = 3.2 x 10-13

    So δE/E = δv/c
     
    Last edited: Jun 4, 2010
  7. Jun 11, 2010 #6
    Hello Bob,
    I really dont get any clue for conversion (it not a home work problem).
    As we know that:
    [tex]
    \tau\;\Gamma\geq\hbar.
    [/tex]
    Here [tex]\tau=141.8169[/tex] ns and [tex]\hbar=6.58211899\times10^{-16}[/tex] eVs.
    So using the above formula i get
    [tex]\Gamma=4.6413 [/tex] neV.
    I really do not get any hint for conversion of the unit of [tex]\Gamma[/tex] to s-1 or s.
    thanks for your help
     
  8. Jun 13, 2010 #7
    Rajini-
    The relationship between the decay time and the natural energy linewidth broadening arises from the Fourier transform from time domain to frequency domain (hence photon energy since E=hω/2π). See the section titled

    "The Lorentzian function is a model of homogeneous broadening"

    in

    http://chsfpc5.chem.ncsu.edu/~franzen/CH795Z/ps/2002/lecture/lecture26/lineshape/lineshape.html [Broken]

    Bob S
     
    Last edited by a moderator: May 4, 2017
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