This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Step 1: Understand the composition of a hydrogen atom and Positronium. A hydrogen atom consists of a proton (mass m_p, charge +e) and an electron (mass m_e, charge -e). Its ground state energy is E_H = -13.6 eV. Positronium consists of a positron (mass m_e, charge +e) and an electron (mass m_e, charge -e). Step 2: Calculate the reduced mass for a hydrogen atom. The reduced mass for a two-particle system with masses m_1 and m_2 is given by = (m_1 m_2)/(m_1 + m_2). For a hydrogen atom, m_1 = m_p and m_2 = m_e. _H = (m_p m_e)/(m_p + m_e) Since the mass of a proton (m_p) is much greater than the mass of an electron (m_e), m_p + m_e ≈ m_p. Therefore, _H ≈ (m_p m_e)/(m_p) = m_e. Step 3: Calculate the reduced mass for Positronium. For Positronium, m_1 = m_e (positron) and m_2 = m_e (electron). _Ps = (m_e · m_e)/(m_e + m_e) = (m_e^2)/(2m_e) = (m_e)/(2) Step 4: Determine the energy of Positronium in its ground state. The energy levels of a hydrogen-like atom are directly proportional to the reduced mass of the system. E Therefore, the ratio of the ground state energy of Positronium to that of hydrogen is: E_PsE_H = _Ps_H Using the approximate values for reduced masses: E_PsE_H = (m_e/2)/(m_e) = (1)/(2) So, E_Ps = (1)/(2) E_H. Given E_H = -13.6 eV for the ground state of hydrogen: E_Ps = (1)/(2) (-13.6 eV) E_Ps = -6.8 eV The more precise calculation using _H = (m_p m_e)/(m_p + m_e) yields: E_PsE_H = (m_e/2)/(m_p m_e)m_p + m_e = (m_e)/(2) · (m_p + m_e)/(m_p m_e) = (m_p + m_e)/(2m_p) Since m_p ≈ 1836 m_e: E_PsE_H = (1836 m_e + m_e)/(2 × 1836 m_e) = (1837 m_e)/(3672 m_e) ≈ 0.50027 This value is very close to 0.5, so the approximation holds well. E_Ps = -13.6 eV × 0.50027 ≈ -6.8036 eV Rounding to one decimal place, this is -6.8 eV. The correct option is (c). (c) -6.8 eV That's 2 down. 3 left today — send the next one.