Question

The ionization energy of hydrogen atom in excited state is:

• (A) 4 eV
• (B) 3.4 eV
• (C) 1.51eV
• (D) 1.9eV

Answer 1

The Correct Option is B

Explanation:
The ionization energy of the hydrogen atom in the excited state means the atom is already in the first excited state and n  = 2.The energy of the electron in any excited state is given by,$E_n=-13.6\frac{Z^2}{n^2}\mathrm{e}\mathrm{V}$The above equation can be written for the excited state of the hydrogen atom as,$E_n=\frac{-13.6}{2^2}\mathrm{e}\mathrm{V}$$=-3.4\mathrm{e}\mathrm{V}$Hence, the correct option is (B).