The energy of the electron in the ground state of H-atom is `-13.6 eV`. The energy of the first excited state will be

Options

A. `-1.51 eV`
B. `-3.4 eV`
C. `+1.51`
D. `+3.eV`

Answer

To find the energy of the first excited state of the hydrogen atom, we can use the formula derived from Bohr's model of the atom. Here’s a step-by-step solution:

### Step 1: Understand the Energy Formula
The energy of an electron in a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
where:
- \( E_n \) is the energy of the electron at the principal quantum number \( n \),
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n \) is the principal quantum number (1 for ground state, 2 for the first excited state).

### Step 2: Identify the Ground State Energy
For the ground state (where \( n = 1 \)):
\[ E_1 = -\frac{13.6 \, \text{eV} \cdot 1^2}{1^2} = -13.6 \, \text{eV} \]

### Step 3: Calculate the Energy for the First Excited State
For the first excited state (where \( n = 2 \)):
\[ E_2 = -\frac{13.6 \, \text{eV} \cdot 1^2}{2^2} \]
\[ E_2 = -\frac{13.6 \, \text{eV}}{4} \]

### Step 4: Simplify the Calculation
Now, divide \( -13.6 \, \text{eV} \) by 4:
\[ E_2 = -3.4 \, \text{eV} \]

### Conclusion
The energy of the first excited state of the hydrogen atom is:
\[ E_2 = -3.4 \, \text{eV} \]

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