The wave number of electromagnetic radiation emitted during the transition of electron in between two levels of `Li^(2+)` ion whose principal quantum number is 4 and difference is 2 is `:`

Options

A. `3.5R`
B. `4R`
C. `8R`
D. `(8)/(9)R`

Answer

To calculate the wave number of electromagnetic radiation emitted during the transition of an electron in the `Li^(2+)` ion, we can follow these steps:

### Step 1: Understand the Given Information
We know:
- The sum of the principal quantum numbers (n1 + n2) = 4
- The difference of the principal quantum numbers (n2 - n1) = 2

### Step 2: Set Up the Equations
From the information provided:
1. n1 + n2 = 4
2. n2 - n1 = 2

### Step 3: Solve the Equations
We can solve these two equations simultaneously.

From equation (2):
- n2 = n1 + 2

Substituting this into equation (1):
- n1 + (n1 + 2) = 4
- 2n1 + 2 = 4
- 2n1 = 2
- n1 = 1

Now substituting n1 back to find n2:
- n2 = n1 + 2 = 1 + 2 = 3

So we have:
- n1 = 1
- n2 = 3

### Step 4: Use the Formula for Wave Number
The formula for the wave number (ṽ) of the emitted radiation is given by:
\[ \tilde{\nu} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Where:
- R = Rydberg constant (approximately \(1.097 \times 10^7 \, \text{m}^{-1}\))
- Z = atomic number of lithium (for `Li^(2+)`, Z = 3)
- n1 = 1
- n2 = 3

### Step 5: Substitute the Values into the Formula
Now substituting the values into the formula:
\[ \tilde{\nu} = R \cdot 3^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) \]
\[ \tilde{\nu} = R \cdot 9 \left( 1 - \frac{1}{9} \right) \]
\[ \tilde{\nu} = R \cdot 9 \left( \frac{8}{9} \right) \]
\[ \tilde{\nu} = 8R \]

### Step 6: Conclusion
Thus, the wave number of the electromagnetic radiation emitted during the transition is \(8R\).