The wave number of first line of Balmer series of hydrogen is `1520m^(-1)`. The wave number of first Balmer line of `Li^(2+)` ion in `m^(-1)` is x. Find the vlaue of `(x)/(100)`.

Answer

To solve the problem, we need to find the wave number of the first line of the Balmer series for the \( \text{Li}^{2+} \) ion and then calculate \( \frac{x}{100} \).

### Step-by-Step Solution

1. **Understanding the Balmer Series**:
The Balmer series corresponds to transitions in hydrogen where an electron falls from a higher energy level (n2) to the second energy level (n1 = 2). The first line of the Balmer series corresponds to the transition from \( n_2 = 3 \) to \( n_1 = 2 \).

2. **Wave Number Formula**:
The wave number (\( \bar{\nu} \)) can be calculated using the formula:
\[ \bar{\nu} = R_H \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R_H \) is the Rydberg constant for hydrogen (\( R_H \approx 1.097 \times 10^7 \, \text{m}^{-1} \)).
- \( Z \) is the atomic number of the ion.
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the energy levels.

3. **Given Data for Hydrogen**:
For hydrogen, we know:
- \( \bar{\nu} = 1520 \, \text{m}^{-1} \)
- \( n_1 = 2 \), \( n_2 = 3 \)
- \( Z = 1 \) (for hydrogen)

Plugging in the values for hydrogen:
\[ 1520 = R_H \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

4. **Calculating the Factor**:
Calculate \( \frac{1}{2^2} - \frac{1}{3^2} \):
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]

5. **Finding Rydberg Constant**:
Substitute back to find \( R_H \):
\[ 1520 = R_H \cdot \frac{5}{36} \]
Rearranging gives:
\[ R_H = 1520 \cdot \frac{36}{5} = 10944 \, \text{m}^{-1} \]

6. **Calculating for \( \text{Li}^{2+} \)**:
For \( \text{Li}^{2+} \), we have:
- \( Z = 3 \)
- Using the same \( n_1 \) and \( n_2 \):
\[ \bar{\nu} = R_H \cdot 3^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
\[ \bar{\nu} = 10944 \cdot 9 \cdot \frac{5}{36} \]

7. **Calculating \( \bar{\nu} \)**:
\[ \bar{\nu} = 10944 \cdot 9 \cdot \frac{5}{36} = 10944 \cdot \frac{45}{36} = 10944 \cdot 1.25 = 13680 \, \text{m}^{-1} \]

8. **Finding \( x \)**:
Thus, \( x = 13680 \, \text{m}^{-1} \).

9. **Calculating \( \frac{x}{100} \)**:
\[ \frac{x}{100} = \frac{13680}{100} = 136.8 \]

### Final Answer:
The value of \( \frac{x}{100} \) is \( 136.8 \).