Calculate the wavelength emitted during the transition of an electron in between two level of `Li^(2+)` ion whose sum is `4` and difference is `2`.

Options

A. `1.14xx10^(-6)m`
B. `2.4xx10^(-6)m`
C. `5.2xx10^(-6)m`
D. `3.11xx10^(-6)m`

Answer

To solve the problem of calculating the wavelength emitted during the transition of an electron in the `Li^(2+)` ion, we will follow these steps:

### Step 1: Identify the quantum levels
We are given that the sum of the two quantum levels \( n_1 \) and \( n_2 \) is 4, and their difference is 2. We can set up the following equations:

\[ n_1 + n_2 = 4 \quad \text{(1)} \]
\[ n_2 - n_1 = 2 \quad \text{(2)} \]

### Step 2: Solve the equations
From equation (2), we can express \( n_2 \) in terms of \( n_1 \):

\[ n_2 = n_1 + 2 \]

Substituting this into equation (1):

\[ n_1 + (n_1 + 2) = 4 \]
\[ 2n_1 + 2 = 4 \]
\[ 2n_1 = 2 \]
\[ n_1 = 1 \]

Now, substituting \( n_1 \) back into equation (1) to find \( n_2 \):

\[ n_2 = 4 - n_1 = 4 - 1 = 3 \]

Thus, we have \( n_1 = 1 \) and \( n_2 = 3 \).

### Step 3: Use the Rydberg formula
The Rydberg formula for the wavelength \( \lambda \) of the emitted photon during the transition is given by:

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

Where:
- \( R_H \) is the Rydberg constant (\( 109678 \, \text{cm}^{-1} \))
- \( Z \) is the atomic number of Lithium (\( Z = 3 \))

### Step 4: Calculate the wavelength
Substituting the values into the formula:

\[ \frac{1}{\lambda} = 109678 \cdot 3^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) \]

Calculating \( 3^2 \):

\[ 3^2 = 9 \]

Now calculating the fractions:

\[ \frac{1}{1^2} - \frac{1}{3^2} = 1 - \frac{1}{9} = \frac{9}{9} - \frac{1}{9} = \frac{8}{9} \]

Now substituting back into the equation:

\[ \frac{1}{\lambda} = 109678 \cdot 9 \cdot \frac{8}{9} \]

The \( 9 \) cancels out:

\[ \frac{1}{\lambda} = 109678 \cdot 8 \]

Calculating this gives:

\[ \frac{1}{\lambda} = 878144 \]

Now, taking the reciprocal to find \( \lambda \):

\[ \lambda = \frac{1}{878144} \, \text{cm} \]

### Step 5: Convert to meters
To convert from centimeters to meters:

\[ \lambda = \frac{1}{878144} \times 10^{-2} \, \text{m} \]

Calculating this gives:

\[ \lambda \approx 1.14 \times 10^{-4} \, \text{m} \]

### Final Answer
The wavelength emitted during the transition is approximately \( 1.14 \times 10^{-4} \, \text{m} \).

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