How Is Doppler Velocity Calculated for Recoil Compensation in Mossbauer Effect?

[Mr LoganC]

Homework Statement

Calculate the Doppler velocity needed to compensate for the recoil energy.

Homework Equations

V=\frac{E_{0}}{m c}

The Attempt at a Solution

I found the recoil energy to be 1.95\times10^{-3} eV.
And for Iron-57, E_{\gamma} is 14.4KeV. Which is approximatly equal to E_{0}.
So just plugging these values into the above equation, I get 81.27m/s which seems a bit high, as many articles site around 10mm/s!

Perhaps units I am using should be in something else? Or maybe I am using the completely wrong equation?

 

[M Quack]

From the wiki: http://en.wikipedia.org/wiki/Mössbauer_effect

The emission and absorption of x-rays by gases had been observed previously, and it was expected that a similar phenomenon would be found for gamma rays, which are created by nuclear transitions (as opposed to x-rays, which are typically produced by electronic transitions). However, attempts to observe gamma-ray resonance in gases failed due to energy being lost to recoil, preventing resonance (the Doppler effect also broadens the gamma-ray spectrum). Mössbauer was able to observe resonance in solid iridium, which raised the question of why gamma-ray resonance was possible in solids, but not in gases. Mössbauer proposed that, for the case of atoms bound into a solid, under certain circumstances a fraction of the nuclear events could occur essentially without recoil. He attributed the observed resonance to this recoil-free fraction of nuclear events.

 

[Mr LoganC]

From the wiki: http://en.wikipedia.org/wiki/Mössbauer_effect

Ohhh... and of course I'm calculating it for a "Free" atom, not one bound in a solid! So the higher velocity makes sense!

Thank you so much for the help!

 

[Rajini]

Homework Statement

Calculate the Doppler velocity needed to compensate for the recoil energy.

Homework Equations

V=\frac{E_{0}}{m c}

The Attempt at a Solution

I found the recoil energy to be 1.95\times10^{-3} eV.
And for Iron-57, E_{\gamma} is 14.4KeV. Which is approximatly equal to E_{0}.
So just plugging these values into the above equation, I get 81.27m/s which seems a bit high, as many articles site around 10mm/s!

Perhaps units I am using should be in something else? Or maybe I am using the completely wrong equation?

Hi,
You answer for recoil velocity seems to be correct (i get 81.47 m/s). The 10 mm/s velocity is not the recoil velocity. It is the speed of transducer, i.e. the vibrating velocity of the Mössbauer source to acquire a complete Mössbauer spectrum..1 mm/s is equal to 48.075 neV (E_{D}=\frac{E_{\gamma} v}{c}[\itex], take v=1 mm/s and you will get 48.075 neV).<br /> Cheers, Rajini