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WB JEE Medical WB JEE Medical Solved Paper-2014

  • question_answer
    The ionization energy of hydrogen is 13.6 eV. The energy of the photon released when an electron jumps from the first excited state \[(n=2)\] to the ground state of a hydrogen atom is

    A)  3.4eV           

    B)  4.53eV

    C)  10.2eV          

    D)  13.6eV

    Correct Answer: C

    Solution :

     Energy required to removed electron in the \[(n=2)\] stage The energy of the photon \[=R\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{n}^{2}}} \right)\Rightarrow 13.6\,eV\left( \frac{1}{1}-\frac{1}{4} \right)\] \[=13.6\,eV\left( \frac{4-1}{4} \right)=13.6\,eV\times \frac{3}{4}=\frac{40.8}{4}eV\] \[=10.2\,eV\]


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