Imdadul haque Asked a Question
January 14, 2021 11:54 ampts 30 pts
(C)n=2, He A doubly ionized lithium atom in an excited state (n = 6) emits a photon of energy 4.25 eV. What are the quantum number (n) and the energy (E) of the final state? (d)n 3, LP (b) n=3, E=-13.6 eV (d)n-5, E-4.90 eV (a)n-2, E=-30.6 eVv (C)n-4, E=-7.65 eV
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  • Suman Kumar
    option C is correct
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  • Dinesh khalmaniya thankyou
    option C
    • cropped5669357847373570943.jpg
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  • Priyanshu kumar Best Answer
    option C
    • cropped6770926922738268751.jpg
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    Priyanshu kumar
    please ask if any doubt
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