The ionisation energy of `H` atom is `21.79xx10^(-19) J`. The the value of binding energy of second excited state of `Li^(2+)` ion

Options

A. `3^(2)xx21.7xx10^(-19)J`
B. `21.79xx10^(-19) J`
C. `1/3xx21.79xx10^(-19) J`
D. `1/3^(2)xx21.79xx10^(-19) J`

Explanation

To find the binding energy of the second excited state of the \( \text{Li}^{2+} \) ion, we can follow these steps:

### Step 1: Understand the Ionization Energy of Hydrogen
The ionization energy of the hydrogen atom is given as:
\[ E_{ionization} = 21.79 \times 10^{-19} \, \text{J} \]
This value corresponds to the energy required to remove an electron from the ground state of the hydrogen atom.

### Step 2: Use the Formula for Binding Energy
The binding energy \( E \) of an electron in a hydrogen-like atom can be calculated using the formula:
\[ E = \frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
where:
- \( Z \) is the atomic number of the element,
- \( n \) is the principal quantum number (energy level).

### Step 3: Convert Ionization Energy from Joules to Electron Volts
We know that:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Thus, we can convert the ionization energy of hydrogen from joules to electron volts:
\[ E_{ionization} = \frac{21.79 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} = 13.61 \, \text{eV} \]
This value is consistent with the known ionization energy of hydrogen.

### Step 4: Determine the Values for Lithium Ion
For \( \text{Li}^{2+} \):
- The atomic number \( Z = 3 \) (since lithium has 3 protons).
- The second excited state corresponds to \( n = 3 \).

### Step 5: Calculate the Binding Energy for \( \text{Li}^{2+} \)
Using the formula for binding energy:
\[ E = \frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
Substituting the values:
\[ E = \frac{13.6 \, \text{eV} \cdot 3^2}{3^2} = \frac{13.6 \, \text{eV} \cdot 9}{9} = 13.6 \, \text{eV} \]

### Step 6: Convert the Binding Energy Back to Joules
Now, we convert the binding energy back to joules:
\[ E = 13.6 \, \text{eV} \cdot 1.6 \times 10^{-19} \, \text{J/eV} = 21.76 \times 10^{-19} \, \text{J} \]

### Final Answer
The binding energy of the second excited state of \( \text{Li}^{2+} \) ion is:
\[ E = 21.76 \times 10^{-19} \, \text{J} \]