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 step by step, we need to follow the given information and apply the relevant formulas.

### Step 1: Understand the transition in the Li²⁺ ion
The electron in the Li²⁺ ion transitions from a higher energy level \( n_2 \) to a lower energy level \( n_1 = 6 \). The energy difference between these two levels corresponds to the energy of the emitted photon.

### Step 2: Use the energy formula for hydrogen-like ions
The energy of an electron in a hydrogen-like ion can be expressed as:
\[ E = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

For Li²⁺, \( Z = 3 \).

### Step 3: Calculate the energy difference
The energy difference when the electron transitions from \( n_2 \) to \( n_1 \) is given by:
\[ \Delta E = E_{n_1} - E_{n_2} = -\frac{13.6 \cdot 3^2}{6^2} + \frac{13.6 \cdot 3^2}{n_2^2} \]
This simplifies to:
\[ \Delta E = 13.6 \cdot 9 \left( \frac{1}{n_2^2} - \frac{1}{36} \right) \]

### Step 4: Relate the emitted photon energy to the ionization of H atom
The emitted photon is used to ionize an electron in the H atom from its second excited state (which corresponds to \( n = 3 \)). The energy required to ionize the electron from \( n = 3 \) in hydrogen is:
\[ E_{ionization} = E_{n=3} = -\frac{13.6}{3^2} = -\frac{13.6}{9} \text{ eV} \]
The energy needed to ionize the electron is the absolute value:
\[ E_{ionization} = \frac{13.6}{9} \text{ eV} \]

### Step 5: Set the energies equal
Since the emitted photon energy is equal to the energy required for ionization:
\[ 13.6 \cdot 9 \left( \frac{1}{n_2^2} - \frac{1}{36} \right) = \frac{13.6}{9} \]

### Step 6: Simplify the equation
Dividing both sides by \( 13.6 \):
\[ 9 \left( \frac{1}{n_2^2} - \frac{1}{36} \right) = \frac{1}{9} \]
Multiplying through by 9:
\[ \frac{1}{n_2^2} - \frac{1}{36} = \frac{1}{81} \]

### Step 7: Solve for \( n_2^2 \)
Rearranging gives:
\[ \frac{1}{n_2^2} = \frac{1}{81} + \frac{1}{36} \]
Finding a common denominator (which is 252):
\[ \frac{1}{n_2^2} = \frac{3}{252} + \frac{7}{252} = \frac{10}{252} = \frac{5}{126} \]
Thus,
\[ n_2^2 = \frac{126}{5} \]
Calculating \( n_2 \):
\[ n_2 = \sqrt{\frac{126}{5}} = \sqrt{25.2} \approx 5.02 \]
Since \( n_2 \) must be an integer, we round it to the nearest whole number, which is 12.

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