# If the binding energy of the electron in a hydrogen  atom is $\dpi{100} 13.6\: eV$ the energy required to remove the electron from the first excited state of $\dpi{100} Li^{++}$ is Option 1) $30.6\: eV$ Option 2) $13.6\: eV$ Option 3) $3.4\: eV$ Option 4) $122.4\: eV$

P perimeter

As we learnt in

Energy emitted due to transition of electron -

$\Delta E= Rhcz^{2}\left ( \frac{1}{n_{f}\, ^{2}}-\frac{1}{n_{i}\, ^{2}} \right )$

$\frac{1}{\lambda }= Rz^{2}\left ( \frac{-1}{n_{i}\, ^{2}}+\frac{1}{n_{f}\, ^{2}} \right )$

- wherein

$R= R hydberg\: constant$

$n_{i}= initial state \\n_{f}= final \: state$

$\Delta E=E_{0}z^{2} \left(\frac{1}{z^{2}}-\frac{1}{\infty^{2}} \right )=\frac{E_{0}z^{2}}{4}$

$\because\ \;z=3\ \; \Rightarrow\ \;\Delta E=\frac{13.6\times9}{4}=30.6\ eV$

Correct option is 1.

Option 1)

$30.6\: eV$

This is the correct option.

Option 2)

$13.6\: eV$

This is an incorrect option.

Option 3)

$3.4\: eV$

This is an incorrect option.

Option 4)

$122.4\: eV$

This is an incorrect option.

Exams
Articles
Questions