GREPhysics.NET: GR0177 #31

GR0177 #31

Problem

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Atomic $\Rightarrow$ }Positronium

The positronium atom ground-state energy can be deduced from recalling the reduced mass. $\mu=m/2$ , since one has a positron and electron orbiting each other, and thus the energy, which is related to mass, is half of that of Hydrogen. $E_{positronium}=-13.6/2 eV = -6.8eV$

The Bohr formula still applies, and thus $E=E_1 \left(1/n_f^2 - 1/n_i^2 \right)=8E_1/9$ , which is choice (A).

Alternate Solutions

FutureDrSteve
2011-11-07 12:22:35

Oops, this was supposed to be a separate post, not a reply....

If you realize that the ground state energy of positronium is roughly -6.8eV (which I think you have to in order to solve this problem), you don't have to do any math. An electron transitioning down from n= $\infty$ would have initial potential energy of 0, and drop down to final energy -6.8eV, for a max energy change of 6.8eV, eliminating (C), (D), and (E). Since the electron is only falling down from n=3, it stands to reason that the photon would be less energetic than 6.8eV, leaving only answer (A).

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bill92
2014-10-27 13:17:41

I just took the exam and there was no positronium question! All my studying was for naught!

Just kidding, I think it went well. Thanks to Yosun for this very helpful site!

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FutureDrSteve
2011-11-07 12:22:35

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Tommy Koulax
2007-10-31 17:48:57

Why is the reduced mass m/2 ?

nick1234
2007-11-02 17:16:11

$\mu = \frac{m_{1} \times m_{2}}{m_{1} + m_{2}}$

For the hydrogen atom, $m_{proton}$ dominates the denominator, and $\mu$ becomes $\frac{m_{e} \times m_{p}}{m_{p}} = m_{e}$

For positronium, $m_{1} = m_{2} = m{e}$ , so $\mu = \frac{m_{e}^{2}}{2 m_{e}} = \frac{m_{e}}{2}$

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rinnie
2007-03-26 21:27:49

Hydrogen levels E3 - E1 = -1.5 + 13.6 = 12.1/2 = 6.05.
Energy level of positronium is half those of Hydrogen.

FutureDrSteve
2011-11-07 12:20:53

If you realize that the ground state energy of positronium is roughly -6.8eV (which I think you have to in order to solve this problem), you don't have to do any math. An electron transitioning down from n= $\infty$ would have initial potential energy of 0, and drop down to final energy -6.8eV, for a max energy change of 6.8eV, eliminating (C), (D), and (E). Since the electron is only falling down from n=3, it stands to reason that the photon would be less energetic than 6.8eV, leaving only answer (A).

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