The wave number of the first line of Balmer series of hydrogen is `15200 cm ^(-1)` The wave number of the first Balmer line of `Li^(2+)` ion is

Options

A. `136800 cm^(-1)`
B. `60800 cm^(-1)`
C. `15200 cm^(-1)`
D. `76000 cm^(-1)`

Answer

To find the wave number of the first line of the Balmer series of the \( \text{Li}^{2+} \) ion, we can use the formula for wave number (\( \bar{V} \)) in the context of the Rydberg formula:

\[ \bar{V} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number,
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the electron transitions.

### Step 1: Identify the values for hydrogen
For the first line of the Balmer series in hydrogen:
- The transition is from \( n_2 = 3 \) to \( n_1 = 2 \).
- The wave number given for hydrogen is \( \bar{V}_{H} = 15200 \, \text{cm}^{-1} \).

### Step 2: Write the equation for hydrogen
Using the Rydberg formula for hydrogen (\( Z = 1 \)):
\[ \bar{V}_{H} = R \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Calculating the term inside the parentheses:
\[ \frac{1}{2^2} - \frac{1}{3^2} = \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
Thus,
\[ \bar{V}_{H} = R \cdot \frac{5}{36} \]

### Step 3: Write the equation for \( \text{Li}^{2+} \)
For \( \text{Li}^{2+} \) (where \( Z = 3 \)):
- The transition is still from \( n_2 = 3 \) to \( n_1 = 2 \):
\[ \bar{V}_{\text{Li}^{2+}} = R \cdot 3^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Substituting the previously calculated term:
\[ \bar{V}_{\text{Li}^{2+}} = R \cdot 9 \cdot \frac{5}{36} \]

### Step 4: Relate the wave numbers
Since we know \( \bar{V}_{H} = R \cdot \frac{5}{36} \):
\[ \bar{V}_{\text{Li}^{2+}} = 9 \cdot \bar{V}_{H} \]
Substituting \( \bar{V}_{H} = 15200 \, \text{cm}^{-1} \):
\[ \bar{V}_{\text{Li}^{2+}} = 9 \cdot 15200 \]

### Step 5: Calculate the wave number for \( \text{Li}^{2+} \)
\[ \bar{V}_{\text{Li}^{2+}} = 136800 \, \text{cm}^{-1} \]

### Final Result
The wave number of the first Balmer line of the \( \text{Li}^{2+} \) ion is \( 136800 \, \text{cm}^{-1} \).

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