If the binding energy of ground state electron in a hydrogen atom is $13.6\, \mathrm{eV}$, then, the energy required to remove the electron from the second excited state of $\mathrm{Li}^{2+}$ will be : $x \times 10^{-1} \mathrm{eV}$. The value of $x$ is ________.
Answer (integer)
136
Solution
$\mathrm{E}_{H}=13.6$
<br/><br/>$$
\begin{aligned}
& \mathrm{E}_{\mathrm{Li}^{2+}}=13.6 \frac{Z^{2}}{n^{2}}=13.6 \times \frac{9}{9}=13.6 \mathrm{eV} \\\\
& =136 \times 10^{-1} \mathrm{eV}
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Bohr's Model of Hydrogen Atom
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