Question

An electron with energy $12.09$ eV strikes hydrogen atom in ground state and gives its all energy to the hydrogen atom. Therefore hydrogen atom is excited to __________ state.

• A. Fourth
• B. Third
• C. Second
• D. First

Answer 1

The correct option is B Third

We know that energy of Hydrogen atom in it's orbit is given by formula $\frac{-13.6}{n^2}eV$.

Initial Energy of Electron = $-13.6eV$ (Given that atom in i's ground state i.e n=1)

Energy after striking = $(-13.6+12.09)eV=-1.51eV$

$\Rightarrow$ $\frac{-13.6}{n^2}eV=1.51eV$ $\Rightarrow$ $n^2=9$ $\Rightarrow$ $n=3.$

So, hydrogen atom is excited to third state.

Therefore, B is correct option.