what is the wavelength (in nm) of the spectral line associated with a transition from n=3 to n=2 for `Li^2`+ ion (R=109677 `cm^-1`)

Options

A. 73.39
B. 102
C. 114
D. 43.14

Answer

To find the wavelength of the spectral line associated with the transition from n=3 to n=2 for the `Li^2+` ion, we will use the Rydberg formula. Here’s a step-by-step solution:

### Step 1: Understand the Rydberg Formula
The Rydberg formula for the wavelength of light emitted during an electronic transition is given by:

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

Where:
- \(\lambda\) is the wavelength,
- \(R_z\) is the Rydberg constant for the ion,
- \(n_1\) and \(n_2\) are the principal quantum numbers of the lower and upper energy levels respectively.

### Step 2: Identify the Variables
For the `Li^2+` ion:
- The atomic number \(Z\) of lithium is 3, so \(R_z = R \cdot Z^2\).
- Given \(R = 109677 \, \text{cm}^{-1}\).
- The transition is from \(n_2 = 3\) to \(n_1 = 2\).

### Step 3: Calculate \(R_z\)
Calculate \(R_z\):

\[ R_z = R \cdot Z^2 = 109677 \cdot 3^2 = 109677 \cdot 9 = 987093 \, \text{cm}^{-1} \]

### Step 4: Plug Values into the Rydberg Formula
Substituting \(n_1\), \(n_2\), and \(R_z\) into the Rydberg formula:

\[ \frac{1}{\lambda} = 987093 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

Calculating the fractions:

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

Finding a common denominator (36):

\[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36} \]

So,

\[ \frac{1}{2^2} - \frac{1}{3^2} = \frac{9}{36} - \frac{4}{36} = \frac{5}{36} \]

### Step 5: Substitute Back to Find \(\lambda\)
Now substituting back:

\[ \frac{1}{\lambda} = 987093 \cdot \frac{5}{36} \]

Calculating:

\[ \frac{1}{\lambda} = \frac{4935465}{36} \approx 137649.03 \, \text{cm}^{-1} \]

### Step 6: Calculate \(\lambda\)
Taking the reciprocal to find \(\lambda\):

\[ \lambda = \frac{1}{137649.03} \approx 7.26 \times 10^{-6} \, \text{cm} \]

### Step 7: Convert to Nanometers
Convert \(\lambda\) from centimeters to nanometers:

\[ \lambda = 7.26 \times 10^{-6} \, \text{cm} \times 10^7 \, \text{nm/cm} = 72.6 \, \text{nm} \]

### Final Answer
The wavelength of the spectral line associated with the transition from \(n=3\) to \(n=2\) for the `Li^2+` ion is approximately **72.6 nm**.

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