The difference between the wave number of 1st line of Balmer series and last line of Paschen series for `Li^(2+)` ion is :

Options

A. `(R)/(36)`
B. `(5R)/(36)`
C. 4R
D. `(R)/(4)`

Answer

To solve the problem of finding the difference between the wave number of the first line of the Balmer series and the last line of the Paschen series for the `Li^(2+)` ion, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Series**:
- The first line of the Balmer series corresponds to the transition from \( n_2 = 3 \) to \( n_1 = 2 \).
- The last line of the Paschen series corresponds to the transition from \( n_2 = \infty \) to \( n_1 = 3 \).

2. **Use the Rydberg Formula**:
The wave number \( \bar{\nu} \) is given by the Rydberg formula:
\[ \bar{\nu} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, and \( Z \) is the atomic number.

3. **Calculate the Wave Number for the Balmer Series**:
For the first line of the Balmer series:
- \( n_1 = 2 \)
- \( n_2 = 3 \)
\[ \bar{\nu}_{\text{Balmer}} = RZ^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = RZ^2 \left( \frac{1}{4} - \frac{1}{9} \right) \]
\[ = RZ^2 \left( \frac{9 - 4}{36} \right) = RZ^2 \left( \frac{5}{36} \right) \]

4. **Calculate the Wave Number for the Paschen Series**:
For the last line of the Paschen series:
- \( n_1 = 3 \)
- \( n_2 = \infty \)
\[ \bar{\nu}_{\text{Paschen}} = RZ^2 \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) = RZ^2 \left( \frac{1}{9} - 0 \right) = RZ^2 \left( \frac{1}{9} \right) \]

5. **Calculate the Difference**:
Now, we need to find the difference between the wave numbers:
\[ \Delta \bar{\nu} = \bar{\nu}_{\text{Balmer}} - \bar{\nu}_{\text{Paschen}} \]
\[ = RZ^2 \left( \frac{5}{36} \right) - RZ^2 \left( \frac{1}{9} \right) \]
To combine these fractions, we need a common denominator:
\[ = RZ^2 \left( \frac{5}{36} - \frac{4}{36} \right) = RZ^2 \left( \frac{1}{36} \right) \]

6. **Substitute the Value of Z**:
For the `Li^(2+)` ion, \( Z = 3 \):
\[ \Delta \bar{\nu} = R \cdot 3^2 \cdot \frac{1}{36} = R \cdot 9 \cdot \frac{1}{36} = \frac{R}{4} \]

### Final Answer:
The difference between the wave number of the first line of the Balmer series and the last line of the Paschen series for the `Li^(2+)` ion is:
\[ \frac{R}{4} \]