Question

Ionisation energy of $Li$ (Lithium) atom in ground state is $5.4eV$. 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. $81.0eV$
• B. $135.4eV$
• C. $203.4eV$
• D. $156.6eV$

Answer 1

Answer: (C)

Given- $\ast$ First ionization energy $=5.4eV$
* second ionization energy $=75.6eV$.
* For third ionization energy-
- we know that energy required for exitation of $e^{-}$ is given as -
$I{E}_2=13.6z^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]eV$
- For $L{i}^{2+}$ ion, we have $z=3$
- For ionization of $e^{-}$from ground state, we have
$\begin{array}{rl}& n_1=1n_2\to \infty \\ \Rightarrow & IE{E}_3=13.6\times (3{)}^2\times \left[\frac{1}{1^2}-\frac{1}{\infty }\right]eV\\ =& 122.4eV\end{array}$
$\Rightarrow$ Energy required to remove all the electron from $Li$ atom is given as
$\begin{array}{rl}{E}_{\text{Total }}& =I{E}_1+I{E}_2+I{E}_3\\ & =5.4eV+75.6eV+122.4eV\\ & =203.4eV\end{array}$