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

JEE Main & Advanced Physics Atomic Physics Question Bank Atomic Structure

  • question_answer
    In hydrogen atom, if the difference in the energy of the electron in \[n=2\] and \[n=3\] orbits is E, the ionization energy of hydrogen atom is                                   [EAMCET (Med.) 2000]

    A)            13.2 E                                      

    B)            7.2 E

    C)            5.6 E                                         

    D)            3.2 E

    Correct Answer: B

    Solution :

                       Energy \[=0.0258\ amu\]      (K = constant)                    n1 = 2 and n2 = 3, so \[E=K\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right]=K\left[ \frac{5}{36} \right]\]                    For removing an electron n1 = 1 to \[{{n}_{2}}=\infty \]                    Energy \[{{E}_{1}}=K[1]=\frac{36}{5}E=7.2\ E\]            \[\therefore \]Ionization energy = 7.2 E


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