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10 February, 22:23

A ground state hydrogen atom absorbs a photon of light having a wavelength of 92.05 nm. It then gives off a photon having a wavelength of 1736 nm. What is the final state of the hydrogen atom?

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  1. 10 February, 23:01
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    Absorbed photon energy

    Ea = hc/λ ... (Planck's equation)

    Ea = hc / 92.05^-9m

    Energy emitted

    Ee = hc / 1736^-9m

    Energy retained ...

    ∆E = Ea - Ee = hc (1/92.05 ^-9 - 1/1736^-9)

    ∆E = (6.625^-34) (3.0^8) (1.028^7)

    ∆E = 2.04^-18 J

    Converting J to eV (1.60^-19 J/eV)

    ∆E = 2.04^-18 / 1.60^-19

    ∆E = 12.70 eV

    Ground state (n=1) energy for Hydrogen = - 13.60eV

    New energy state = (-13.60 + 12.70) eV = - 0.85 eV

    Energy states for Hydrogen

    En = - (13.60 / n²)

    n² = - 13.60 / - 0.85 = 16

    n = 4
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