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 \( x \) for the first line of the Balmer series of the \( Li^{2+} \) ion and then compute \( \frac{x}{100} \).

### Step-by-Step Solution:

1. **Understanding the Formula for Wave Number**:
The wave number \( \bar{\nu} \) for a hydrogen-like atom can be expressed using the formula:
\[ \bar{\nu} = z^2 \cdot R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( z \) = atomic number
- \( R \) = Rydberg constant (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \))
- \( n_1 \) = lower energy level (for the first line of the Balmer series, \( n_1 = 2 \))
- \( n_2 \) = higher energy level (for the first line of the Balmer series, \( n_2 = 3 \))

2. **Wave Number for Hydrogen**:
For hydrogen, the wave number of the first line of the Balmer series is given as:
\[ \bar{\nu}_{H} = 1520 \, \text{m}^{-1} \]

3. **Calculating Wave Number for \( Li^{2+} \)**:
For the \( Li^{2+} \) ion, the atomic number \( z = 3 \). Using the same formula:
\[ \bar{\nu}_{Li^{2+}} = z^2 \cdot R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Substituting \( z = 3 \), \( n_1 = 2 \), and \( n_2 = 3 \):
\[ \bar{\nu}_{Li^{2+}} = 3^2 \cdot R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Simplifying this:
\[ \bar{\nu}_{Li^{2+}} = 9 \cdot R \left( \frac{1}{4} - \frac{1}{9} \right) \]

4. **Finding the Common Denominator**:
To subtract the fractions:
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
Thus:
\[ \bar{\nu}_{Li^{2+}} = 9 \cdot R \cdot \frac{5}{36} \]

5. **Substituting the Value of R**:
Using \( R = 1.097 \times 10^7 \, \text{m}^{-1} \):
\[ \bar{\nu}_{Li^{2+}} = 9 \cdot (1.097 \times 10^7) \cdot \frac{5}{36} \]

6. **Calculating \( \bar{\nu}_{Li^{2+}} \)**:
\[ \bar{\nu}_{Li^{2+}} = \frac{9 \cdot 5 \cdot 1.097 \times 10^7}{36} \]
\[ = \frac{49.365 \times 10^7}{36} \approx 1.371 \times 10^7 \, \text{m}^{-1} \]

7. **Final Calculation**:
Now, we need to find \( \frac{x}{100} \):
\[ x = 1.371 \times 10^7 \, \text{m}^{-1} \]
\[ \frac{x}{100} = \frac{1.371 \times 10^7}{100} = 1.371 \times 10^5 \, \text{m}^{-1} \]

### Final Answer:
The value of \( \frac{x}{100} \) is approximately \( 137.1 \).