Ground state energy of H-atom is -13.6 eV. The energy needed to ionise H-atom from its second excited state is

Options

A. 1.51 eV
B. 3.4 eV
C. 13.6 eV
D. 12.1 eV

Answer

To find the energy needed to ionize a hydrogen atom from its second excited state, we can follow these steps:

### Step 1: Understand the Energy Levels of Hydrogen Atom
The energy levels of a hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

### Step 2: Identify the Second Excited State
The second excited state corresponds to \( n = 3 \) (the ground state is \( n = 1 \), the first excited state is \( n = 2 \), and the second excited state is \( n = 3 \)).

### Step 3: Calculate the Energy of the Second Excited State
Using the formula for energy levels:
\[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} \approx -1.51 \, \text{eV} \]

### Step 4: Calculate the Energy Needed for Ionization
To ionize the atom from the second excited state, we need to bring the energy from \( E_3 \) to 0 eV (the energy of a free electron). The energy needed for ionization is:
\[ \text{Energy needed} = 0 \, \text{eV} - E_3 = 0 \, \text{eV} - (-1.51 \, \text{eV}) = 1.51 \, \text{eV} \]

### Conclusion
The energy needed to ionize the hydrogen atom from its second excited state is approximately **1.51 eV**.

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