Atomic acceleration from Photon emission ?

[Widdekind]

The energy-time [uncertainty] relation is usually interpreted, in a practical sense, to signify that the lifetime of an emission (the amount of time taken for the intensity of the light emitted to decay to some separate proportion of its initial intensity) will be uncertain by an amount related to the uncertainty in its energy. The uncertainty in the lifetime can be translated into an uncertainty in the exact moment of emission of a quantum particle. In other words, the more sharply we can measure (in time) the lifetime, and, hence, the moment of creation, of a quantum particle... the more uncertain will be its energy, and vice versa.

 

Jim Baggott. Beyond Measure, pg. 38.

 

 

The wave train produced by a single quantum "transition" is demonstrably one or two feet long [1-2 ns]; th time an atom requires to radiate this wave train is about the same as the lifetime of an excited state.

 

W.H.Cropper. The Quantum Physicists, pg. 99.

 

 

The photon picture of light suggests, that the process of light emission is rather like firing a bullet from a gun, while that of light absorption is similar to a bullet hitting a target. This image correctly predicts, that an atom emitting or absorbing a photon reacts by recoiling.

 

Hey & Walters. The New Quantum Universe, pg. 144.

 

 

Math:

 

For the atom, of mass m, emitting a photon of energy E = h f, w.h.t.:

 

[math]a = \frac{F}{m} \approx \frac{p}{\Delta t} = \frac{ h f }{c \, \Delta t}[/math]

 

[math]\therefore a \approx \lambda_{C} \frac{f}{\Delta t}[/math]

For a hydrogen atom [math]( \lambda_{C} \approx 1.5 fm )[/math], emitting a 3 eV photon, the acceleration is about 10⁸ Gs (terran standard). is this correct ?

[swansont]

Math:

 

For the atom, of mass m, emitting a photon of energy E = h f, w.h.t.:

 

[math]a = \frac{F}{m} \approx \frac{p}{\Delta t} = \frac{ h f }{c \, \Delta t}[/math]

 

[math]\therefore a \approx \lambda_{C} \frac{f}{\Delta t}[/math]

For a hydrogen atom [math]( \lambda_{C} \approx 1.5 fm )[/math], emitting a 3 eV photon, the acceleration is about 10⁸ Gs (terran standard). is this correct ?

 

I think you meant

[math]a = \frac{F}{m} \approx \frac{p}{m\Delta t} = \frac{ h f }{m c \, \Delta t}[/math]

 

 

What is the value of [math]{\Delta t}?[/math]

 

It's supposed to be the lifetime of the excited state.

[Widdekind]

I think you meant

[math]a = \frac{F}{m} \approx \frac{p}{m\Delta t} = \frac{ h f }{m c \, \Delta t}[/math]

 

 

What is the value of [math]{\Delta t}?[/math]

 

It's supposed to be the lifetime of the excited state.

 

Yes, thanks for the correction

[Bob_for_short]

So what is the question? Of course, an atom recoils while radiating.

[Widdekind]

I wanted to double check -- matter must be rather rugged, if it can commonly accommodate hundreds of millions of Gs (!!). I guess that's why the can put payloads, in surface super-gun shells, and sling the same into space.

[Widdekind]

I think you meant

[math]a = \frac{F}{m} \approx \frac{p}{m\Delta t} = \frac{ h f }{m c \, \Delta t}[/math]

 

 

What is the value of [math]{\Delta t}?[/math]

 

It's supposed to be the lifetime of the excited state.

 

Thus,

 

[math]a \approx \frac{p}{m\Delta t} = \frac{ E_{transition} }{m c \, \Delta t} \approx \frac{ E_{transition} \, \Delta E_{state}}{m c \hbar / 2}[/math]

If so, then 'sharp' excited states, with well defined energy values (small [math]\Delta E[/math]), which last longer, would accelerate less. Yet, 'intuitively', why wouldn't more sharply defined energy states, suffer 'sharper' transitions, back down to the ground state, and so accelerate more on emission, even if emission was a long time in coming (lower overall avg. accel.)?

[swansont]

You have to know over what time period the emission is happening in order to get an acceleration. A long lifetime does not mean the decay always occurs after a long time, just that it is likely to. But if you check to see if the system is still in the excited state, you reset the clock; you can dramatically increase the time spent in the excited state this way. (Quantum Zeno effect)