The wavelength (in Å) of an emission line obtained for `Li^(2+)` during an electronic transition `n_(2) = 2` and n = 1 is (R = Rydberg constant)

Options

A. `(3R)/(4)`
B. `(27R)/(4)`
C. `(4)/(3R)`
D. `(4)/(27R)`

Answer

To find the wavelength of the emission line for the `Li^(2+)` ion during the electronic transition from `n = 2` to `n = 1`, we can use the Rydberg formula for hydrogen-like atoms. Here’s a step-by-step solution:

### Step 1: Identify the Rydberg Formula
The Rydberg formula for the wavelength of light emitted during an electronic transition in a hydrogen-like atom is given by:

\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

where:
- \(\lambda\) is the wavelength,
- \(R\) is the Rydberg constant,
- \(Z\) is the atomic number,
- \(n_1\) is the lower energy level,
- \(n_2\) is the higher energy level.

### Step 2: Substitute the Values
For `Li^(2+)`, the atomic number \(Z\) is 3 (since lithium has an atomic number of 3). The transition is from \(n_2 = 2\) to \(n_1 = 1\).

Substituting these values into the formula:

\[ \frac{1}{\lambda} = R \cdot 3^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]

### Step 3: Calculate the Terms
Now calculate \(3^2\) and the fractions:

\[ 3^2 = 9 \]
\[ \frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{4}{4} - \frac{1}{4} = \frac{3}{4} \]

### Step 4: Combine the Results
Now substitute these results back into the equation:

\[ \frac{1}{\lambda} = R \cdot 9 \cdot \frac{3}{4} \]
\[ \frac{1}{\lambda} = \frac{27R}{4} \]

### Step 5: Solve for Wavelength
To find \(\lambda\), take the reciprocal:

\[ \lambda = \frac{4}{27R} \]

### Final Result
Thus, the wavelength of the emission line for the transition from \(n = 2\) to \(n = 1\) in `Li^(2+)` is:

\[ \lambda = \frac{4}{27R} \]