Question

Ionization energy of $Li$(Lithium) atom in ground state in $5.4eV$. Binding energy of an electron in $L{i}^{+}$ ion in ground state is $75.6eV$. Energy required to remove all three electrons of Lithium (Li) atom is:-

• A. $203.4eV$
• B. $135.4eV$
• C. $81.0eV$
• D. $156.6eV$

Answer 1

The correct option is A $203.4eV$

$\begin{array}{c}\text{Given-}\ast \text{First ionization energy}=5.4eV\\ \text{* second ionization energy}=75.6eV\text{.}\\ \text{* For third ionization energy-}\\ \text{- we know that energy required for exitation of}{e}^{-}\\ \text{is given as -}\\ I{E}_2=13.6z^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]eV\end{array}$

$\begin{array}{c}{\text{- For Li}}^{2+}\text{ion, we have}z=3\\ \text{- For ionization of}{e}^{-}\text{from ground state, we have}\\ n_1=1n_2\to \infty \\ \Rightarrow I{E}_3=13.6\times (3{)}^2\times [\frac{1}{1^2}-\frac{1}{\infty }]eV\\ =122.4eV\end{array}$

$\begin{array}{c}\Rightarrow \text{Energy required to remove all the electron from Li atom}\\ \text{is given as}\\ \begin{array}{cc}{E}_{\text{Total}}& =I{E}_1+I{E}_2+I{E}_3\\ & =5.4ev+75.6eV+122.4eV\\ & =203.4eV\end{array}\\ \text{Hence, opt}\left(a\right)\text{is the correct answer.}\end{array}$