STD 11 - 2. structure of atom — NEET STD 11 Science — Question

For the electron in ground state i.e. $\mathrm{n}=1,$ the ionization energy is $I E_{1}=E_{\infty}-E_{1}$

$I E_{1}=0-(-13.6)=$ (since, ground state energy is $\left.(-13.6 \mathrm{eV})\right)$

$I E_{1}=13.6 \mathrm{eV}$

For the electron in first excited state i.e. $\mathrm{n}=2$, the ionization energy is $I E_{2}=\frac{I E_{1}}{n^{2}}$

$I E_{2}=\frac{I E_{1}}{2^{2}}$

$I E_{2}=\frac{13.6}{4}=3.4 \mathrm{eV}$

For the electron in second excited state i.e. $\mathrm{n}=3$, the ionization energy is $I E_{2}=\frac{I E_{1}}{n^{2}}$

$I E_{2}=\frac{I E_{1}}{3^{2}}$

$I E_{2}=\frac{13.6}{9}=1.51 \mathrm{eV}$ and so on.

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