Question

An absorption of $12.088eV$ energy by an electron in ground state of $H$-atom brings in the excitation of electron to which orbit?

• A. Two
• B. Three
• C. Four
• D. Not possible. Minimum energy required is -13.6eV

Answer 1

The correct option is B Three

The energy absorbed is 12.088 eV. It is given by the expression $E=-13.6\times (\frac{1}{n_2^2}-\frac{1}{n_1^2})eV$Substitute values in the above expression.
$12.088=-13.6\times (\frac{1}{n_1^2}-\frac{1}{1^2})$
Hence, $n_2=3$. Thus, the electron is excited to third orbit.