A hydrogen atom,initially at rest in its grou | vedclass

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(B) $1$. Energy of the incident photon $h v_1$ is used to ionize the hydrogen atom and provide kinetic energy to the electron:$h v_1 = E_{ionization}

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(B) $1$. Energy of the incident photon $h v_1$ is used to ionize the hydrogen atom and provide kinetic energy to the electron:$h v_1 = E_{ionization} + KE_{electron} = 13.6 eV + 10 eV = 23.6 eV$.$2$. The positronium atom is a system of an electron and a positron. Its reduced mass $\mu$ is given by $\mu = \frac{m_e m_p}{m_e + m_p} = \frac{m_e^2}{2m_e} = \frac{m_e}{2}$.$3$. The ground state energy of the positronium atom is $E_p = -13.6 eV 	imes \frac{\mu}{m_e} = -13.6 eV 	imes \frac{1}{2} = -6.8 eV$.$4$. When the electron (with $10 eV$ kinetic energy) combines with a stationary positron to form a positronium atom,the total energy available is $10 eV$. This energy is used to form the positronium atom in its ground state (releasing $6.8 eV$ as a photon) and to provide kinetic energy to the center of mass of the positronium atom $(5 eV)$.$5$. By conservation of energy: $10 eV = h v_2 + |E_p| + KE_{COM} = h v_2 + 6.8 eV + 5 eV$.$h v_2 = 10 eV - 11.8 eV = -1.8 eV$. Wait,re-evaluating: The energy released as a photon $h v_2$ is the difference between the initial kinetic energy of the electron and the sum of the binding energy and the kinetic energy of the center of mass of the positronium atom.$h v_2 = KE_{electron} - (|E_p| + KE_{COM}) = 10 eV - (6.8 eV + 5 eV) = -1.8 eV$. Actually,the energy balance is: $KE_{electron} + |E_p| = h v_2 + KE_{COM}$.$10 eV + 6.8 eV = h v_2 + 5 eV \implies h v_2 = 11.8 eV$.$6$. The difference between the two photon energies is $|h v_1 - h v_2| = |23.6 eV - 11.8 eV| = 11.8 eV$.

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