An electron in a hydrogen atom absorbs a photon and jumps to ahigher orbit. a) f

Question

An electron in a hydrogen atom absorbs a photon and jumps to ahigher orbit. a) find the energy the photon must have if the initial sate isn=3 and the final state is n=5. b) find the frequency of the photon that find in part a. c) if the intial state were n=5 and the final state were n=7,would the energy of the photon be greater than, less than r equalto the amount you found in part a? Explain. d) calculate the photon energy for part C, to verify youranswer. a) find the energy the photon must have if the initial sate isn=3 and the final state is n=5. b) find the frequency of the photon that find in part a. c) if the intial state were n=5 and the final state were n=7,would the energy of the photon be greater than, less than r equalto the amount you found in part a? Explain. d) calculate the photon energy for part C, to verify youranswer.

Explanation / Answer

    Energy levels of electron  En = (-13.6 e V ) Z2 / n2 energy of the electron at n = 3state                E3   = -13.6 /32                       = -1.51 eV energy of the electron at n = 5state                E5   = -13.6 /52                       = -0.544 eV energy of photon           E =  E3   -E5              = -1.51 -0.544   = -0.966 eV frequency of photon           =E / h              =    0.966 / (4.1468*10-15 eV .s)             = 2.32*1014 Hz if the intial state were n=5 and the final state weren=7,            energy of the electron at n = 5state                E5   = -13.6 /52                       = -0.544 eV energy of the electron at n = 5state                E7   = -13.6 /72                       = -0.277eV energy of photon           E =  E3   -E5              = -0.544+0.277   = -0.2664 eV energy of photn is less than the value of parta              E5   = -13.6 /52                       = -0.544 eV energy of photon           E =  E3   -E5              = -1.51 -0.544   = -0.966 eV frequency of photon           =E / h              =    0.966 / (4.1468*10-15 eV .s)             = 2.32*1014 Hz if the intial state were n=5 and the final state weren=7,            energy of the electron at n = 5state                E5   = -13.6 /52                       = -0.544 eV energy of the electron at n = 5state                E7   = -13.6 /72                       = -0.277eV energy of photon           E =  E3   -E5              = -0.544+0.277   = -0.2664 eV energy of photn is less than the value of parta              E5   = -13.6 /52                       = -0.544 eV energy of the electron at n = 5state                E7   = -13.6 /72                       = -0.277eV energy of photon           E =  E3   -E5              = -0.544+0.277   = -0.2664 eV energy of photn is less than the value of parta              E7   = -13.6 /72                       = -0.277eV energy of photon           E =  E3   -E5              = -0.544+0.277   = -0.2664 eV energy of photn is less than the value of parta

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