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 the hydrogen atom from its second excited state, we can follow these steps:

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

### Step 2: Calculate the Energy of the Second Excited State
The second excited state corresponds to \( n = 3 \). We can calculate the energy at this level:
\[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} = -1.51 \, \text{eV} \]

### Step 3: Determine the Energy Required for Ionization
Ionization energy is the energy required to remove the electron from the atom, which means we need to bring the energy of the electron from its current state to zero (the energy level at infinity).

To ionize the electron from the second excited state (which has an energy of -1.51 eV), we need to supply enough energy to raise it to 0 eV:
\[ \text{Ionization Energy} = 0 \, \text{eV} - (-1.51 \, \text{eV}) = 1.51 \, \text{eV} \]

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

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