Which of the following is the energy of a possible excited state of hydrogen?

Since, at n = 1, the population of electrons is maximum i.e. at ground state. So, maximum excitation will take place from n = 1 to n= 2

Hence, n= 2 is the possible excited state. Now, we have the formula for energy of

  H – atom   ${\left(E_n\right)}_H=-13.6\frac{Z^2}{n^2}eV$

Where, Z = atomic number, Z for H – atom = 1

$\therefore {\left(E_n\right)}_H=-13.6\times \frac{1}{2^2}eV$

$=-\frac{13.6}{4}eV=-3.4eV$