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JEE Main & Advanced Physics Atomic Physics Question Bank Self Evaluation Test - Atoms

  • question_answer
    If radiation corresponding to first line of "Balmer series" of \[H{{e}^{+}}\] ion knocked out electron from 1st excited state of H atom, the kinetic energy of ejected electron from H atom would be (eV) - [Given \[{{E}_{n}}=-\frac{{{Z}^{2}}}{{{n}^{2}}}\left( 13.6eV \right)\]]

    A)  4.155 eV

    B)  8.310 eV

    C)  2.515 eV

    D)  5.550 eV

    Correct Answer: A

    Solution :

    [a] Energy of photon corresponding to first line of Balmer series \[=\left( 13.6 \right){{\left( 2 \right)}^{2}}\left[ \frac{1}{4}-\frac{1}{9} \right]\] Energy need to eject electron from n=2 level in H atom \[=\left( 13.6 \right)\left( \frac{1}{4} \right)\] So, required kinetic energy \[=\left( 13.6 \right)\left[ \left( \frac{1}{4}-\frac{1}{9} \right)-\left( \frac{1}{4} \right) \right]eV\]\[=13.6\times \left( \frac{11}{36} \right)=4.155eV\]


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