A singly ionized helium atom in an excited state (n=4) emits a photon of energy 2.6eV. Given that the ground state energy of hydrogen atom is ?13.6eV, the energy (Ef?) and quantum number (n) of the resulting state are respectively,
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Step 1
Given: Initial energy of helium ion (Ei) = -3.4 eV Energy of the emitted photon (ΔE) = 2.6 eV Energy of the resulting state (Ef) = Ei - ΔE Ef = -3.4 eV - 2.6 eV Ef = -6.0 eV Show more…
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