Convert Linewidth \Gamma of Nuclear Level to Seconds/Inverse Seconds

[Rajini]

Hello all,

is there any possibility to express line width \Gamma of the nuclear level in second (s) or inverse second ?
For eg. for 57Fe,
\Gamma=4.6413 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

 

[Bob S]

Use Planck's constant

h/2π = 6.582118 x 10^-16 eV-seconds

Bob S

 

[Meir Achuz]

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.

 

[Rajini]

hello,
i think i have not supplied some information.
In a book i found the following relation:
\Gamma/\omega_N<1. Here \Gamma is nuclear line width and \omega_N is Larmor precession frequency.
But i prefer to write as \tau_N<(1/\Gamma)---is this correct?[\tau_N=(\omega_N)^{-1}]
What i understand is the unit of \tau_N (obviously unit is s) and \Gamma should be same. Also \Gamma can take unit as mm/s, eV, etc and \omega_N may be in Hz i guess. But we also know that inverse of \omega_N 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

 

[Bob S]

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

 

[Rajini]

Hello Bob,
I really don't get any clue for conversion (it not a home work problem).
As we know that:
<br /> \tau\;\Gamma\geq\hbar.<br />
Here \tau=141.8169 ns and \hbar=6.58211899\times10^{-16} eVs.
So using the above formula i get
\Gamma=4.6413 neV.
I really do not get any hint for conversion of the unit of \Gamma to s^-1 or s.
thanks for your help

 

[Bob S]

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

Bob S