yxgao
Dec16-04, 01:27 AM
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.
E_n = -\frac{\mu}{2{\hbar}^2} (\frac{Ze^2}{4\pi\epsilon_0})^2 \frac{1}{n^2}
For hydrgen, the effective mass is approximately the mass of the electron.
\mu=m_e
However, for positronium,
\mu=\frac{m_pm_e}{m_p+m_e}=\frac{m_em_e}{m_e+m_e}= \frac{m_e}{2}
So the length of the energy is only half that of the hydrogen atom.
Therefore,
E_3-E_1=(\frac{1}{3^2}-1)(-6.8 eV)=-\frac{8}{9}*-6.8 eV = 6.04 eV
grn. 31.
Edit:
Nevermind, I figured it out.
E_n = -\frac{\mu}{2{\hbar}^2} (\frac{Ze^2}{4\pi\epsilon_0})^2 \frac{1}{n^2}
For hydrgen, the effective mass is approximately the mass of the electron.
\mu=m_e
However, for positronium,
\mu=\frac{m_pm_e}{m_p+m_e}=\frac{m_em_e}{m_e+m_e}= \frac{m_e}{2}
So the length of the energy is only half that of the hydrogen atom.
Therefore,
E_3-E_1=(\frac{1}{3^2}-1)(-6.8 eV)=-\frac{8}{9}*-6.8 eV = 6.04 eV
grn. 31.