wave number of the first line of Paschen series in `Be^(3+)` ion is :

Options

A. `(7R)/16`
B. `(7R)/144`
C. `(7R)/9`
D. `R/144`

Answer

To find the wave number of the first line of the Paschen series in the \( \text{Be}^{3+} \) ion, we can follow these steps:

### Step 1: Understand the Rydberg Formula
The wave number \( \bar{\mu} \) (in cm\(^{-1}\)) is given by the Rydberg formula:
\[ \bar{\mu} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R \) is the Rydberg constant (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( Z \) is the atomic number,
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the energy levels involved in the transition.

### Step 2: Identify the Values
For the \( \text{Be}^{3+} \) ion:
- The atomic number \( Z = 4 \) (since beryllium has an atomic number of 4).
- For the first line of the Paschen series, the transitions occur from \( n_2 = 4 \) to \( n_1 = 3 \).

### Step 3: Substitute the Values into the Formula
Now, substitute \( Z = 4 \), \( n_1 = 3 \), and \( n_2 = 4 \) into the Rydberg formula:
\[ \bar{\mu} = R \cdot 4^2 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \]

### Step 4: Calculate the Terms
Calculate \( 4^2 = 16 \), \( \frac{1}{3^2} = \frac{1}{9} \), and \( \frac{1}{4^2} = \frac{1}{16} \):
\[ \bar{\mu} = R \cdot 16 \left( \frac{1}{9} - \frac{1}{16} \right) \]

### Step 5: Find a Common Denominator
To subtract the fractions \( \frac{1}{9} \) and \( \frac{1}{16} \), find a common denominator:
- The least common multiple of 9 and 16 is 144.
- Convert \( \frac{1}{9} \) to \( \frac{16}{144} \) and \( \frac{1}{16} \) to \( \frac{9}{144} \):
\[ \frac{1}{9} - \frac{1}{16} = \frac{16}{144} - \frac{9}{144} = \frac{7}{144} \]

### Step 6: Substitute Back into the Wave Number Equation
Now substitute back into the wave number equation:
\[ \bar{\mu} = R \cdot 16 \cdot \frac{7}{144} \]
\[ \bar{\mu} = \frac{112R}{144} = \frac{7R}{9} \]

### Step 7: Final Result
Thus, the wave number of the first line of the Paschen series in the \( \text{Be}^{3+} \) ion is:
\[ \bar{\mu} = \frac{7R}{9} \]