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(B) Positronium is a binary system consisting of an electron and a positron. Both particles revolve around their common center of mass.To treat this a

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-$6.8$ eV

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(B) Positronium is a binary system consisting of an electron and a positron. Both particles revolve around their common center of mass.To treat this as a single-particle system,we use the reduced mass $\mu$ of the binary system:$\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{m_e 	imes m_e}{m_e + m_e} = \frac{m_e}{2} = \frac{m}{2}$The ground state energy of an electron in a hydrogen atom is given by:$E_1 = -\frac{m e^4}{8 \epsilon_0^2 h^2} = -13.6 	ext{ eV}$Since the energy of the ground state is directly proportional to the mass of the orbiting particle,the ground state energy of positronium $E_1'$ is:$E_1' = \frac{\mu}{m} E_1 = \frac{m/2}{m} E_1 = \frac{1}{2} E_1$Substituting the value of $E_1$:$E_1' = \frac{1}{2} (-13.6 	ext{ eV}) = -6.8 	ext{ eV}$

Positronium is just like a $H$-atom with the proton replaced by the positively charged antiparticle of the electron (called the positron,which is as massive as the electron). What would be the ground state energy of positronium?

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