# Linewidth and time

1. Jun 4, 2010

### 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

2. Jun 4, 2010

### Bob S

Use Planck's constant

h/2π = 6.582118 x 10-16 eV-seconds

Bob S

3. Jun 4, 2010

### 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.

Last edited: Jun 4, 2010
4. Jun 4, 2010

### 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

5. Jun 4, 2010

### 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

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
6. Jun 11, 2010

### Rajini

Hello Bob,
I really dont get any clue for conversion (it not a home work problem).
As we know that:
$$\tau\;\Gamma\geq\hbar.$$
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.

7. Jun 13, 2010

### 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 [Broken]

Bob S

Last edited by a moderator: May 4, 2017