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Wave Optics

Question
CBSEENPH12040090

Ionization potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. According to Bohr's theory, the spectral lines emitted by hydrogen will be

  • two 

  • three

  • four

  • one

Solution

B.

three

Ionization energy corresponding to ionization potential  = -13.6 eV.
Photon energy incident = 12.1 eV
So, the energy of electron in excited state
    = -13.6 + 12.1 = -1.5 eV
i.e.,      straight E subscript straight n space equals space minus fraction numerator 13.6 over denominator straight n squared end fraction space rightwards double arrow space space straight n squared space equals space minus fraction numerator 13.6 over denominator negative 1.5 end fraction space equals space 9
therefore space space space space space space straight n space equals space 3
i.e., energy of electron in excited state corresponds to third orbit.
The possible spectral lines are when electron jumps from orbit 3rd to 2nd; 3rd to 1st and 2nd to 1st. Thus, 3 spectral lines are emitted.