The wave number of the first Balmer line `Li^(2+)` ion is `136800 cm^(-1)` .The wave number of the first line of Balmer series of hydrogen atom is` ("in" cm^(-1))`

Options

A. 68400
B. 15200
C. 76000
D. 30800

Explanation

To find the wave number of the first line of the Balmer series of the hydrogen atom, we can use the relationship between the wave numbers of the Balmer series for the lithium ion \( Li^{2+} \) and hydrogen.

### Step-by-Step Solution:

1. **Understand the Formula for Wave Number:**
The wave number \( \bar{\nu} \) is given by 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, \( Z \) is the atomic number, and \( n_1 \) and \( n_2 \) are the principal quantum numbers.

2. **Identify the Given Values:**
For the \( Li^{2+} \) ion, we know:
- The wave number \( \bar{\nu}_{Li^{2+}} = 136800 \, \text{cm}^{-1} \)
- The atomic number \( Z \) for lithium \( (Li) \) is 3, so \( Z^2 = 3^2 = 9 \).
- The transition for the first line of the Balmer series corresponds to \( n_1 = 2 \) and \( n_2 = 3 \).

3. **Set Up the Equation for \( Li^{2+} \):**
For the \( Li^{2+} \) ion:
\[ 136800 = R_H \cdot 9 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Calculate \( \frac{1}{2^2} - \frac{1}{3^2} \):
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
Thus, we can rewrite the equation:
\[ 136800 = R_H \cdot 9 \cdot \frac{5}{36} \]

4. **Solve for \( R_H \):**
Rearranging gives:
\[ R_H = \frac{136800 \cdot 36}{9 \cdot 5} \]
Calculate \( R_H \):
\[ R_H = \frac{136800 \cdot 36}{45} = \frac{4924800}{45} = 109440 \, \text{cm}^{-1} \]

5. **Set Up the Equation for Hydrogen:**
For hydrogen, \( Z = 1 \):
\[ \bar{\nu}_{H} = R_H \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
We already calculated \( \frac{1}{2^2} - \frac{1}{3^2} = \frac{5}{36} \), so:
\[ \bar{\nu}_{H} = 109440 \cdot \frac{5}{36} \]

6. **Calculate \( \bar{\nu}_{H} \):**
\[ \bar{\nu}_{H} = \frac{109440 \cdot 5}{36} = \frac{547200}{36} = 15200 \, \text{cm}^{-1} \]

### Final Answer:
The wave number of the first line of the Balmer series of the hydrogen atom is \( 15200 \, \text{cm}^{-1} \).