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**Hint:**For this we need to know the formula of energy of the hydrogen atom given by Bohr’s model. According to the formula, the energy of a hydrogen atom can never be positive. To calculate ground state energy put \[n\] equals to 1 and for the first excited state put \[n{\text{ = 2}}\]and so on.

Formula used: \[{{\text{E}}_{\text{n}}} = - \dfrac{1}{{{{\text{n}}^{\text{2}}}}} \times 13.6{\text{ eV}}\]

Where, \[{{\text{E}}_{\text{n}}}\]is energy of \[{{\text{n}}^{{\text{th}}}}\] state, \[{\text{n}}\] is the number of stationary state\[(1,2,3......)\].

**Complete step by step solution:**

The above question is based on the Bohr model of Hydrogen atom. It assumes that electrons revolve in certain stationary orbits and the whole value of energy is quantized or fixed. Each energy level has fixed energy and the electron exactly needs to have that amount of energy if it wants to circulate in that particular orbit. The energy of hydrogen atom is given by the formula:

\[{{\text{E}}_{\text{n}}} = - \dfrac{1}{{{{\text{n}}^{\text{2}}}}} \times 13.6{\text{ eV}}\]

As we know \[n\] represents a natural number and hence can never be negative. So clearly the energy of each of the states will be negative. So option A and D stands eliminated here. Now let us put the value of \[n\] one by one and see the values:

Putting \[n{\text{ = 1}}\] we will get,

\[{{\text{E}}_{\text{n}}} = - \dfrac{1}{{{1^2}}} \times 13.6{\text{ eV = - 13}}{\text{.6 eV}}\]

For \[n{\text{ = 2}}\]

\[{{\text{E}}_{\text{n}}} = - \dfrac{1}{{{2^2}}} \times 13.6{\text{ eV = - 3}}{\text{.4 eV}}\]

**Hence, the correct option is option C.**

**Note:**We do not start from \[n{\text{ = 0}}\] to calculate ground state of hydrogen, because in that case the value of ground state energy will be infinite. And this is not possible according to Zero point theorem which states that the lowest possible energy of an element should be finite. Hence, for hydrogen we always start with \[n{\text{ = 1}}\]

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