If the bonding energy of the electron in a hydrogen atom is 13.6 eV , then energy required to remove electron from first excited state of $\mathrm{Li}^{2+}$ is

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Aicharya · Accepted Answer

$B . E=13.6 \mathrm{eV}=-E_1 \Rightarrow E_1=-13.6 \mathrm{eV}$ For $\mathrm{Li}^{2+}, \mathrm{z}=3, \mathrm{n}_1=2, \mathrm{n}_2=\infty$ $	herefore \quad \Delta \mathrm{E}=\mathrm{E}_{\infty}-\mathrm{E}_2=0-\left(-13.6 	imes \frac{\mathrm{Z}^2}{\mathrm{~n}_1^2}ight)$ $=13.6 	imes \frac{(3)^2}{(2)^2}=30.6 \mathrm{eV}$

If the bonding energy of the electron in a hydrogen atom is 13.6 eV , then energy required to remove…

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