2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main - Mock Test - 41

  • question_answer
    The binding energy of the electron in the lowest orbit of the hydrogen atom is 13.6 ev. The energies required in eV to remove an electron from three lowest orbits of the hydrogen are-

    A) 13.6, 6.8, 8.4 eV

    B)        13.6, 10.2, 3.4 eV

    C) 13.6, 27.2, 40.8 eV        

    D) 13.6, 3.4, 1.5 eV

    Correct Answer: D

    Solution :

    [d] \[{{E}_{T}}=-13.6\frac{{{Z}^{2}}}{{{n}^{2}}}\] I orbit for (H) \[Z=1,n=1\] \[{{E}_{T}}=-13.6eV\] II orbit for (H) \[Z=1,n=2\] \[{{E}_{T}}=-\frac{13.6}{4}=-3.4eV\] III orbit for (H) \[{{E}_{T}}=-\frac{13.6}{9}=-1.51\,eV\]


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