GREPhysics.NET: GR9677 #12

GR9677 #12

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Atomic $\Rightarrow$ }Positronium

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 proportional to $\mu$ . 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

zwarner0891
2017-08-25 14:34:42

Just let the energy equal the mass of the electron which is equal to the mass of the positron. Then using the reduced mass equation with energy of 13.6 replacing the mass you will get the correct answer

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FutureDrSteve
2011-11-01 13:24:27

I never even heard of positronium during my undergrad studies, but after going through two practice tests already, this was an easy one. Other commenters have mentioned that you can pretty much count on seeing a positronium question on test day. It's worth memorizing that its reduced mass is $\frac{1}{2} m _{e}$ and that its ground state energy is half of Hydrogen's.

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Steve
2011-08-24 16:35:00

Can anyone explain how the reduced mass comes into play and introduces the factor of $\frac{1}{2}$ ?

I began with

$\frac{1}{\lambda}=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$
and
$E=\frac{hc}{\lambda}$
so
$E=\frac{hc}{\lambda}=hcR\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$

Solving this for $n_f=1$ and $n_i=\infty$ yields the characteristic $E=13.6$ eV

Obviously I'm missing the factor of $\frac{1}{2}$ that is supposed to be introduced with the reduced mass somehow?

itorsics
2011-09-11 00:27:05

The formula for the Rydberg constant $R$ is $R = \frac{m_e e^4}{8 \epsilon_0^2 h^3 c}$ .
For positronium, you need to replace the electron mass $m_e$ by the reduced mass.

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Can anyone explain how the reduced mass comes into play and introduces the factor of $\frac{1}{2}$ ? I began with $\frac{1}{\lambda}=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$ and $E=\frac{hc}{\lambda}$ so $E=\frac{hc}{\lambda}=hcR\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$ Solving this for $n_f=1$ and $n_i=\infty$ yields the characteristic $E=13.6$ eV Obviously I'm missing the factor of $\frac{1}{2}$ that is supposed to be introduced with the reduced mass somehow?

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