The energy of a Hydrogen atom in its first excited state is

The correct answer is Option (1) → - 3.4 eV

For a hydrogen atom, the energy of the electron in the nth orbit is given by:

$E_n = -\frac{13.6}{n^2}\ \text{eV}$

For the first excited state, $n = 2$

$E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4\ \text{eV}$

Answer: $E = -3.4\ \text{eV}$