2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. Atomic Physics

JEE Main & Advanced Physics Atomic Physics Question Bank Atomic Structure

  • 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 [CBSE PMT 1992; DPMT 2000; RPMT 1999;  AMU (Med.) 2001]

    A)            3.40 eV                                   

    B)            1.51 eV

    C)            0.85 eV                                   

    D)            0.66 eV

    Correct Answer: D

    Solution :

                       \[{{E}_{3}}=-\frac{13.6}{9}=-1.51\ eV;\] \[{{E}_{4}}=-\frac{13.6}{16}=-0.85\ eV\]            \[\therefore {{E}_{4}}-{{E}_{3}}=0.66\ eV\]


You need to login to perform this action.
You will be redirected in 3 sec spinner