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 find 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 1: Identify the transitions for the Balmer and Paschen 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 \( n_1 = 3 \).

### Step 2: Write the formula for wave number
The wave number \( \bar{\nu} \) can be calculated using the formula:
\[ \bar{\nu} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number (for \( Li^{2+} \), \( Z = 3 \)),
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the electron transitions.

### Step 3: Calculate the wave number for the first line of the Balmer series
For the first line of the Balmer series:
- \( n_1 = 2 \)
- \( n_2 = 3 \)

Substituting these values into the formula:
\[ \bar{\nu}_{\text{Balmer}} = R \cdot 3^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Calculating the fractions:
\[ \bar{\nu}_{\text{Balmer}} = R \cdot 9 \left( \frac{1}{4} - \frac{1}{9} \right) \]
Finding a common denominator (36):
\[ \bar{\nu}_{\text{Balmer}} = R \cdot 9 \left( \frac{9 - 4}{36} \right) = R \cdot 9 \cdot \frac{5}{36} = \frac{45R}{36} = \frac{5R}{4} \]

### Step 4: Calculate the wave number for the last line of the Paschen series
For the last line of the Paschen series:
- \( n_1 = 3 \)
- \( n_2 = \infty \)

Substituting these values into the formula:
\[ \bar{\nu}_{\text{Paschen}} = R \cdot 3^2 \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) \]
Since \( \frac{1}{\infty^2} = 0 \):
\[ \bar{\nu}_{\text{Paschen}} = R \cdot 9 \left( \frac{1}{9} - 0 \right) = R \cdot 9 \cdot \frac{1}{9} = R \]

### Step 5: Find the difference between the two wave numbers
Now, we can find the difference:
\[ \Delta \bar{\nu} = \bar{\nu}_{\text{Balmer}} - \bar{\nu}_{\text{Paschen}} = \frac{5R}{4} - R \]
Converting \( R \) to a fraction with a common denominator:
\[ \Delta \bar{\nu} = \frac{5R}{4} - \frac{4R}{4} = \frac{1R}{4} = \frac{R}{4} \]

### 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:
\[ \Delta \bar{\nu} = \frac{R}{4} \]