Question

The ionisation energy for excited hydrogen atom in $eV$ will be

• A. $13.6$
• B. less than $\mathrm{13.6.}$
• C. greater than $13.6$
• D. $3.4$ or less

Answer 1

The correct option is D $3.4$ or less

The ionization energy (IE) of an atom describes the minimum amount of energy required to remove an electron (to infinity) from the atom.

For the electron in ground state i.e. n = 1, the ionization energy is
$I{E}_1=E_{\infty }-E_1$
$I{E}_1=0-(-13.6)$ .................(since, ground state energy is (-13.6 eV))
$I{E}_1=13.6eV$

For the electron in first excited state i.e. n = 2, the ionization energy is
$I{E}_2=\frac{I{E}_1}{n^2}$

$I{E}_2=\frac{I{E}_1}{2^2}$

$I{E}_2=\frac{13.6}{4}=3.4eV$

For the electron in second excited state i.e. n = 3, the ionization energy is
$I{E}_2=\frac{I{E}_1}{n^2}$

$I{E}_2=\frac{I{E}_1}{3^2}$

$I{E}_2=\frac{13.6}{9}=1.51eV$ and so on.

Hence, the ionisation energy for excited hydrogen atom will be $3.4eV$ or less than it.