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(B) The Rydberg constant $R$ is proportional to the reduced mass $\mu$ of the system.For a hydrogen-like atom,the reduced mass is $\mu = \frac{m_e M}{

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$R_\infty /2$

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(B) The Rydberg constant $R$ is proportional to the reduced mass $\mu$ of the system.For a hydrogen-like atom,the reduced mass is $\mu = \frac{m_e M}{m_e + M}$,where $M$ is the mass of the nucleus. Since $M \gg m_e$,$\mu \approx m_e$,and $R_\infty = \frac{m_e e^4}{8 \epsilon_0^2 h^3 c}$.For positronium,the system consists of an electron $(m_e)$ and a positron $(m_p = m_e)$.The reduced mass $\mu$ for positronium is:$\mu = \frac{m_e 	imes m_p}{m_e + m_p} = \frac{m_e 	imes m_e}{m_e + m_e} = \frac{m_e^2}{2m_e} = \frac{m_e}{2}$.Since $R \propto \mu$,the Rydberg constant for positronium $(R_{pos})$ is related to the Rydberg constant for an infinite nuclear mass $(R_\infty)$ as:$R_{pos} = \frac{\mu}{m_e} R_\infty = \frac{m_e / 2}{m_e} R_\infty = \frac{R_\infty}{2}$.

Positronium consists of an electron and a positron (a particle which has the same mass as an electron,but opposite charge) orbiting round their common centre of mass. Calculate the value of the Rydberg constant for this system.

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