Question

Ionisation energy of $Li$ (Lithium) atom in ground state is $5.4\, eV$. Binding energy of an electron in $Li^+$ ion in ground state is $75.6\, eV$. Energy required to remove all three electrons of Lithium (Li) atom is

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

Answer 1

Answer: (C)

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