Question

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

• A. $30.6\text{eV}$
• B. $13.6\text{eV}$
• C. $3.4\text{eV}$
• D. $122.4\text{eV}$

Answer 1

The correct option is A $30.6\text{eV}$

The energy of the electron in $n^{th}$ shell of a hydrogen like atom is given by

$E_n=\frac{Z^2}{n^2}(-13.6\text{eV})$

For ${\text{Li}}^{2+}$, $Z=3$ and for first excited state $n=2$

$\Rightarrow E=\frac{3^2}{2^2}\times (-13.6\text{eV})$

$\Rightarrow E=-30.6\text{eV}$

The negative sign shows that the electron is bound in the atom with this much energy.

So, an energy of $E=+30.6\text{eV}$ is required to remove an electron from the first excited state of ${\text{Li}}^{2+}$.

Hence, option $\left(A\right)$ is correct.