An electron in `Li^(2+)` ion makes a transition from higher state `n_(2)` to lower state `n_(1)=6.` The emitted photons is used to ionize an electron in H-atom from 2nd excited state. The electron on leaving the H-atom has a de Broglie wavelength `lambda-12.016 "Å"`.Find the value of `n_(2).`
Note : Use `(12.016)^(2)= (150xx144)/(13.6xx11),lambda_("Å")=sqrt((150)/(KE_(eV)))`

Answer

To solve the problem, we need to find the value of \( n_2 \) for the transition of an electron in the \( Li^{2+} \) ion, given that it transitions from a higher energy state \( n_2 \) to a lower energy state \( n_1 = 6 \). The emitted photon is used to ionize an electron in a hydrogen atom from its second excited state, and the electron has a de Broglie wavelength of \( \lambda = 12.016 \, \text{Å} \).

### Step-by-step Solution:

1. **Calculate the Kinetic Energy of the Electron in Hydrogen:**
The de Broglie wavelength \( \lambda \) is related to the kinetic energy \( KE \) of the electron by the formula:
\[ \lambda = \sqrt{\frac{150}{KE}} \]
Rearranging gives:
\[ KE = \frac{150}{\lambda^2} \]
Substituting \( \lambda = 12.016 \, \text{Å} = 12.016 \times 10^{-10} \, \text{m} \):
\[ KE = \frac{150}{(12.016)^2} \]

2. **Calculate \( \lambda^2 \):**
\[ \lambda^2 = (12.016)^2 = 144.384256 \, \text{Å}^2 \]

3. **Calculate the Kinetic Energy:**
\[ KE = \frac{150}{144.384256} \approx 1.039 \, \text{eV} \]

4. **Calculate the Binding Energy of the Electron in the Second Excited State of Hydrogen:**
The second excited state corresponds to \( n = 3 \):
\[ BE = \frac{13.6}{n^2} = \frac{13.6}{3^2} = \frac{13.6}{9} \approx 1.511 \, \text{eV} \]

5. **Total Energy of the Electron:**
The total energy \( E \) of the electron is the sum of its kinetic energy and binding energy:
\[ E = KE + BE = 1.039 + 1.511 \approx 2.550 \, \text{eV} \]

6. **Using the Energy Formula for \( Li^{2+} \):**
The total energy can also be expressed in terms of the energy levels of the lithium ion:
\[ E = 13.6 \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For \( Li^{2+} \), \( Z = 3 \) and \( n_1 = 6 \):
\[ 2.550 = 13.6 \cdot 3^2 \left( \frac{1}{6^2} - \frac{1}{n_2^2} \right) \]
Simplifying gives:
\[ 2.550 = 13.6 \cdot 9 \left( \frac{1}{36} - \frac{1}{n_2^2} \right) \]
\[ 2.550 = 122.4 \left( \frac{1}{36} - \frac{1}{n_2^2} \right) \]

7. **Solving for \( n_2 \):**
Rearranging the equation:
\[ \frac{1}{36} - \frac{1}{n_2^2} = \frac{2.550}{122.4} \]
\[ \frac{1}{36} - \frac{1}{n_2^2} \approx 0.0208 \]
\[ \frac{1}{n_2^2} = \frac{1}{36} - 0.0208 \]
\[ \frac{1}{n_2^2} \approx 0.0278 \]
\[ n_2^2 \approx \frac{1}{0.0278} \approx 36.0 \]
Taking the square root gives:
\[ n_2 \approx 12 \]

### Final Answer:
The value of \( n_2 \) is \( 12 \).