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JEE Main & Advanced Sample Paper JEE Main Sample Paper-4

  • question_answer
    The ionisation energy of Hydrogen atom is 13.6 eV. Following Bohr's theory the energy corresponding to a transition between the 3rd and the 4th orbit is

    A)  3.40eV                                

    B)  1.51eV  

    C)  0.85 eV                               

    D)  0.66 eV

    Correct Answer: D

    Solution :

    \[{{E}_{3}}=-\frac{13.6}{{{3}^{2}}}=-1.51eV\] \[{{E}_{4}}=-\frac{13.6}{{{4}^{2}}}=-0.85eV\] \[\Delta E={{E}_{4}}-{{E}_{3}}=-0.85-(-1.51)\]\[=0.66eV\]


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