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

Explanation

To solve the problem of finding the wave number of the first line of the Balmer series for the \(Li^{2+}\) ion, we can follow these steps:

### Step 1: Understand the Wave Number Formula
The wave number (\( \mu \)) for a transition in a hydrogen-like atom can be expressed using the formula:
\[ \mu = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R \) is the Rydberg constant (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \) or \( 109677 \, \text{cm}^{-1} \)),
- \( Z \) is the atomic number,
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final states, respectively.

### Step 2: Identify the Transition for the Balmer Series
For the first line of the Balmer series, the transition occurs from \( n_2 = 3 \) to \( n_1 = 2 \).

### Step 3: Calculate for Hydrogen
We know the wave number for the first line of the Balmer series for hydrogen (\( H \)) is given as:
\[ \mu_H = 15200 \, \text{cm}^{-1} \]
Using the formula for hydrogen (\( Z = 1 \)):
\[ \mu_H = R \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Calculating the term in the parentheses:
\[ \frac{1}{2^2} - \frac{1}{3^2} = \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
So,
\[ \mu_H = R \cdot \frac{5}{36} \]

### Step 4: Calculate for \( Li^{2+} \)
For the \( Li^{2+} \) ion, \( Z = 3 \). The wave number can be calculated as:
\[ \mu_{Li^{2+}} = R \cdot 3^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Substituting the value of \( Z \):
\[ \mu_{Li^{2+}} = R \cdot 9 \cdot \frac{5}{36} \]
Now substituting \( R \) with the known value:
\[ \mu_{Li^{2+}} = 15200 \cdot 9 \]
Calculating this:
\[ \mu_{Li^{2+}} = 136800 \, \text{cm}^{-1} \]

### Step 5: Conclusion
Thus, the wave number of the first Balmer line of the \( Li^{2+} \) ion is:
\[ \mu_{Li^{2+}} = 136800 \, \text{cm}^{-1} \]