`Li^(2+)` is initially is ground state. When radiation of wavelength `lamda_(0)` incident on it, it emits 6 different wavelengths during de excitation find `lamda_(0)`

Options

A. 11.4 nm
B. 9.4 nm
C. 12.3 nm
D. 10.8 nm

Answer

To solve the problem, we need to determine the wavelength of radiation (λ₀) that causes the lithium ion (Li²⁺) to emit six different wavelengths during de-excitation from its ground state.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We have a lithium ion in the ground state (n=1).
- When radiation of wavelength λ₀ is incident on it, it emits six different wavelengths during de-excitation.

2. **Using the Formula for Number of Emitted Wavelengths:**
- The number of different wavelengths emitted during transitions between energy levels can be calculated using the formula:
\[ N = \frac{n(n-1)}{2} \]
- Here, N is the number of emitted wavelengths, and n is the number of energy levels involved in the transitions.

3. **Setting Up the Equation:**
- Given that N = 6, we can set up the equation:
\[ 6 = \frac{n(n-1)}{2} \]
- Multiplying both sides by 2 gives:
\[ 12 = n(n-1) \]

4. **Finding n:**
- Rearranging the equation gives:
\[ n^2 - n - 12 = 0 \]
- This is a quadratic equation. We can solve it using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Here, a = 1, b = -1, and c = -12:
\[ n = \frac{1 \pm \sqrt{1 + 48}}{2} = \frac{1 \pm 7}{2} \]
- This gives us two possible solutions:
\[ n = 4 \quad \text{(since n must be positive)} \]

5. **Calculating the Wavelength:**
- The transitions occur from n=4 to n=1 (ground state). We can use the Rydberg formula to find the wavelength of the emitted radiation:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- For Li²⁺, Z = 3 (atomic number of lithium):
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \cdot 3^2 \left( \frac{1}{1^2} - \frac{1}{4^2} \right) \]
- Simplifying this:
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \cdot 9 \left( 1 - \frac{1}{16} \right) \]
\[ = 1.097 \times 10^7 \cdot 9 \cdot \frac{15}{16} \]
\[ = 1.097 \times 10^7 \cdot \frac{135}{16} \]
\[ = 1.097 \times 10^7 \cdot 8.4375 \approx 9.27 \times 10^7 \]

6. **Finding λ:**
- Now, taking the reciprocal to find λ:
\[ \lambda \approx \frac{1}{9.27 \times 10^7} \approx 1.08 \times 10^{-8} \text{ m} = 10.8 \text{ nm} \]

### Final Answer:
The wavelength λ₀ is approximately **10.8 nm**.