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## Main Question or Discussion Point

What is the energy of the photon emitted when a positronium atom goes from the n=3 state to the n=1 state?

Edit:

Nevermind, I figured it out.

[tex]

E_n = -\frac{\mu}{2{\hbar}^2} (\frac{Ze^2}{4\pi\epsilon_0})^2 \frac{1}{n^2}

[/tex]

For hydrgen, the effective mass is approximately the mass of the electron.

[tex]\mu=m_e[/tex]

However, for positronium,

[tex]\mu=\frac{m_pm_e}{m_p+m_e}=\frac{m_em_e}{m_e+m_e}=\frac{m_e}{2}[/tex]

So the length of the energy is only half that of the hydrogen atom.

Therefore,

[tex]E_3-E_1=(\frac{1}{3^2}-1)(-6.8 eV)=-\frac{8}{9}*-6.8 eV = 6.04 eV[/tex]

grn. 31.

Edit:

Nevermind, I figured it out.

[tex]

E_n = -\frac{\mu}{2{\hbar}^2} (\frac{Ze^2}{4\pi\epsilon_0})^2 \frac{1}{n^2}

[/tex]

For hydrgen, the effective mass is approximately the mass of the electron.

[tex]\mu=m_e[/tex]

However, for positronium,

[tex]\mu=\frac{m_pm_e}{m_p+m_e}=\frac{m_em_e}{m_e+m_e}=\frac{m_e}{2}[/tex]

So the length of the energy is only half that of the hydrogen atom.

Therefore,

[tex]E_3-E_1=(\frac{1}{3^2}-1)(-6.8 eV)=-\frac{8}{9}*-6.8 eV = 6.04 eV[/tex]

grn. 31.

Last edited: