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Haryana PMT Haryana PMT Solved Paper-2008

  • question_answer
    lonization 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 Bohrs         theory, the spectral lines emitted by hydrogen will be

    A)  two                                      

    B)  three

    C)  four                                     

    D)  one

    Correct Answer: B

    Solution :

                    onization energy corresponding to ionization potential \[=-13.6\text{ }eV\] Photon energy incident \[=12.1\text{ }eV\] So, the energy of electron in excited state                 \[=-13.6+12.1\]                 \[=-1.5eV\] i.e.,        \[{{E}_{n}}=-\frac{-13.6}{{{n}^{2}}}eV\]                 \[-1.5=\frac{-13.6}{{{n}^{2}}}\] \[\Rightarrow \]               \[{{n}^{2}}=\frac{-13.6}{-1.5}\approx 9\] \[\therefore \]  \[n=3\] ie, 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.


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