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Any flaw in this experiment to determine the momentum vector of the electron in a hydrogen atom?

+ 0 like - 2 dislike
87 views

Postulates - -

We know that particles only exhibit wave-like characteristics only when in motion, and also that the electron can be accurately described as a Standing wave, therefore the electron must be in motion around the nucleus.
Since at any one time the electron has a 95% probability to be anywhere in a certain orbital, the electron must be moving at a very high velocity $v$.

Conditions -

A hydrogen atom is suspended at at distance of more than 1 meter from a heavy nucleus which has a sufficient atomic mass such that under its influence pair production can occur.

Experiment -

A photon of energy $E$ $\ge$ $1.022 MeV$ moves past a heavy nucleus, and undergoes pair production. Now the energy of the incident photon would be divided equally into a positron and electron as Follows $$E = KEe + KEp + RMe + RMp -----\alpha$$ where - $KEe$ = Kinetic energy of Electron - $KEp$ = Kinetic energy of Positron - $RMe$ = Rest mass of Electron - $RMp$ = Rest mass of Positron Now by the law of conservation of momentum $$ (\frac{E}{c}) = Pe + Pp -----\beta$$ where $(\frac{E}{c})$ = momentum of the photon $Pe$ = momentum of electron $Pp$ = momentum of positron Now a hydrogen atom is so placed near this pair production site that the positron attacks it. Now by the postulate above the positron collides with the electron and annihilation occurs and we get two photons as a result $$ e^- + e^+ \rightarrow \gamma + \gamma$$ Now the photons will have energies $ E_1 and E_2 $ respectively. and thus will have momenta $ P_1 and P_2 $ respectively. Now again by the law of conservation of momentum: $$ Pp + Pe_H = P_1 + P_2 ---- \omega$$ where $Pe_H$ is the momentum of the electron of the hydrogen atom. Now, from equation $\beta$ and $\omega$ $$Pe_h = P_1 + P_2 + P_e - (\frac{E}{c}) -----\epsilon$$ Now $P_1 = (\frac{E_1}{c})$ and $P_2 = (\frac{E_2}{c})$ replacing, $\epsilon$ becomes: $$Pe_h = (\frac{E_1}{c}) + (\frac{E_2}{c}) + P_e - (\frac{E}{c}) ----\upsilon$$ From $\alpha$ further substitutions and simplifications can be made. As each of the momenta in $\upsilon$ can be measured therefore we can get the momentum of the electron in the hydrogen atom without any uncertainty whatsoever. Knowing the momentum the velocity, and hence the direction of the electron motion can be determined. The point of collision can also be determined with accuracy by finding the point of intersection of the respective trajectories of the incident positron and the trajectories of the photons produced as a result of annihilation. Is there any flaw in this theoretical experiment? Would like to know how plausible it is in its execution and would be in its results

Closed as per community consensus as the post is not graduate-level
asked Feb 18 in Closed Questions by batwayne (-15 points) [ revision history ]
recategorized Feb 18 by Dilaton

not graduate+ level. Users with 500+ reputation may vote here to get it closed.

very implausible, since it is based on assumptions that are gross simplifications of the real situation. You cannot simply postulate whatever you like!





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