Atoms & Nuclei

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201 of 239

Question 201 of 239

If the binding energy of ground state electron in a hydrogen atom is 13.6eV13.6eV, then, the energy required to remove the electron from the second excited state of Li2+Li{}^{2+} will be: x×10−1eVx\times {10}^{-1}eV. The value of xx is _____.

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Answer: 136

Given:
Binding energy, EH=E0=13.6eV{E}_{H}={E}_{0}=13.6eV.
Now for hydrogen like atom, we can write
ELi2+=E0Z2n2=13.6×99=13.6eV{E}_{{Li}^{2+}}={E}_{0}\dfrac{{Z}^{2}}{{n}^{2}} =13.6\times \dfrac{9}{9} =13.6eV
=136×10−1eV=136\times {10}^{-1}eV
Therefore, x=136x=136.

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