Helium atom is excited to a state with the configuration (2s2p) with an energy 58.3 eV. After some time, this atom spontaneously ejects a single electron. The value of the orbital angular momentum quantum number (I) of the ejected electron in the final state of the system is (lonization potential of He(1s)² is 24.6 eV)

Correct option is A

Given:

• Helium atom is excited to the configuration (2s2p) with energy 58.3 eV.
• Ionization potential of ground state He(1s)² is 24.6 eV.
• The atom spontaneously ejects a single electron.
• Determine the orbital angular momentum quantum number $l$ of the ejected electron.

Solution:

1. Understand the excitation state:

   • In the configuration (2s2p), the two electrons occupy different orbitals.
   • The 2s orbital has $l=0$, and the 2p orbital has $l=1$.
   • One of these electrons is ejected when the atom transitions to a final state.

2. Energy considerations:

   • Total energy of the system is 58.3 eV.
   • Ionization potential of He(1s)² is 24.6 eV.
   • When the system loses one electron, the remaining electron will be left in an orbital consistent with the given energy.

3. Identify the ejected electron:

   • The ejected electron comes from the higher energy orbital (2p) because it has a higher angular momentum and energy compared to the 2s orbital.

4. Orbital angular momentum quantum number of the ejected electron:

   • For the 2p orbital, the quantum number $l=1$.
   • Thus, the ejected electron has an angular momentum quantum number $l=1$.

Conclusion:
The orbital angular momentum quantum number $l$ of the ejected electron is (a) 1.