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  GR8677 #99
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Verbatim question for GR8677 #99
Atomic}Positronium

Interestingly, this is almost an exact repeat of Exam 9677 Prob 12. The reasoning's the same:

The positronium atom involves a positron-electron combination instead of the usual proton-electron combo for the H atom. Charge remains the same, and thus one can approximate its eigenvalue by changing the mass of the Rydberg energy (recall that the ground state of the Hydrogen atom is 1 Rydberg).

Recall the reduced mass \mu=\frac{m_1 m_2}{m_1+m_2}, where for identical masses, one obtains \mu = m/2. The Rydberg in the regular Hydrogen energy eigenvalue formula E=R\left(1/n_f^2 - 1/n_i^2 \right) is R\propto \mu, where \mu \rightarrow \mu/2. Substitute in the new value of the reduced mass to get E\approx R/2. R=-13.6 eV, and thus E\approx -6.8 eV.

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Comments
Richard
2007-11-01 13:56:01
I think there is a positronium question akin to this one on all of the released phyiscs GRE's.
Hint I think.
NEC
Blake7
2007-09-26 01:23:16
Yosun,

Don't you mean that 9677.12 is almost an exact repeat of 8677.99 ?

; )
NEC

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I think there is a positronium question akin to this one on all of the released phyiscs GRE's. Hint I think.

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