Question

If \(r_p\) and \(r_H\) are the radius and \(E_p\) and \(E_H\) are the energy of an electron in the \(n\) orbit of positronium atom and hydrogen atom, respectively, then:

• A. \(r_p = 2r_H\) and \(E_p = E_H/2\)
• B. \(r_p = 2r_H\) and \(E_p = 2E_H\)
• C. \(r_p = 2r_H\) and \(E_p = E_H/4\)
• D. \(r_p = r_H\) and \(E_p = 2E_H\)

Hint:

Positronium has reduced mass \(m_e/2\); radius scales as \(1/\mu\) and energy as \(\mu\).

Answer 1

The Correct Option is A

Step 1: The Bohr model with a finite-mass nucleus depends on the reduced mass \(\mu = \dfrac{m_1 m_2}{m_1 + m_2}\). For hydrogen the proton is very heavy, so \(\mu_H \approx m_e\).

Step 2: Positronium is an electron bound to a positron, both of mass \(m_e\). Its reduced mass is \[\mu_p = \frac{m_e \cdot m_e}{m_e + m_e} = \frac{m_e}{2}.\]

Step 3: The Bohr radius scales as \(r_n \propto \dfrac{1}{\mu}\). Since \(\mu_p = \tfrac{1}{2}\mu_H\), the positronium orbit is twice as large: \[r_p = 2\,r_H.\]

Step 4: The binding energy scales as \(E_n \propto \mu\). Since \(\mu_p = \tfrac{1}{2}\mu_H\), the positronium energy is half that of hydrogen in magnitude: \[E_p = \frac{E_H}{2}.\]

Step 5: Combining both results: \[\boxed{r_p = 2r_H,\quad E_p = \frac{E_H}{2}}\]