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JEE Main & Advanced JEE Main Paper (Held On 9 April 2016)

  • question_answer
    A hydrogen atom makes a transition from n = 2 to n=1 and emits a photon. This photon strikes a doubly ionized lithium atom (z = 3) in excited state and completely removes the orbiting electron. The least quantum number for the excited stated of the ion for the process is :   JEE Main Online Paper (Held On 09 April 2016)

    A) 4             

    B) 5

    C) 2                                             

    D) 3

    Correct Answer: A

    Solution :

                    Energy of proton = 13.6 . 3.4 = 10.2eV For removal of electron \[10.2eV>13.6\frac{{{z}^{2}}}{{{n}^{2}}}\] \[{{n}^{2}}>13.6\frac{9}{10.2}\] so minimum value of n = 4                                


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