Answer:
The energetic order third excited > seconds excited > first excited > ground
Explanation:
Bohr's atomic model includes an integer that is responsible for the quantization of energy
E = - 13.6 /n² eV
Where n is the so-called principal quantum number, it is an integer with values of
n = 1, 2, 3… inf
For n = 1 have the so-called ground state that is the one with the lowest energy, for the minus sign in the formula E1 = -13.6 eV
For n = 2 it is the first excited state has the most energy than the ground state,
E2 = 13.6 / 4 = -3.4eV
For n = 3 we have the second excited state, E3 = -13.6 / 9 = -1.5 eV
For n = 4 we have the third excited state, E4 = -0.85 eV
The energetic order of the states is
E4 > E3 > E2 >E1