Energy of H-atom in the ground state is -13.6 eV, hence energy in the second excited state is

Options

A. `-6.8eV`
B. `-3.4eV`
C. `-1.51eV`
D. `-4.53eV`

Answer

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

### Step 1: Understand the given information
The energy of the hydrogen atom in the ground state (n=1) is given as -13.6 eV. The atomic number (Z) of hydrogen is 1.

### Step 2: Identify the formula for energy levels
According to Bohr's model, the energy of an electron in a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, Z^2}{n^2} \, \text{eV} \]
where \(E_n\) is the energy at the principal quantum number \(n\).

### Step 3: Determine the principal quantum number for the second excited state
The ground state corresponds to \(n=1\). The first excited state corresponds to \(n=2\), and the second excited state corresponds to \(n=3\).

### Step 4: Substitute the values into the formula
For the second excited state (n=3):
\[ E_3 = -\frac{13.6 \times 1^2}{3^2} \]

### Step 5: Calculate the energy
Now, calculate \(E_3\):
\[ E_3 = -\frac{13.6}{9} \, \text{eV} \]
\[ E_3 = -1.51 \, \text{eV} \]

### Step 6: Conclusion
The energy of the hydrogen atom in the second excited state is -1.51 eV.

### Final Answer
The energy in the second excited state is -1.51 eV.

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