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QJEEM23Ph153c: JEE Main 2023 Physics - Class 12 JEEM23Ph153c Solution: If the binding energy of ground state electron in a hydrogen…

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

Step-by-Step Verified Solution

EH=13.6\mathrm{E}_{H}=13.6

amp;ELi2+=13.6Z2n2=13.6×99=13.6eVamp;=136×10−1eV\begin{aligned}& \mathrm{E}_{\mathrm{Li}^{2+}}=13.6 \frac{Z^{2}}{n^{2}}=13.6 \times \frac{9}{9}=13.6 \mathrm{eV} \\\\& =136 \times 10^{-1} \mathrm{eV}\end{aligned}

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