The atomic spectrum of `Li^(+2)` ion arises due to the transition of an electron from `n_(2)` to `n_(1)` if `n_(1) +n_(2)=4` and `(n_(2)-n_(1))` =2 then the wavelength of `3^(rd)` line of this series in `Li^(+2)` ion will be

Options

A. 1.08 nm
B. 10.8 nm
C. 108 nm
D. 1080 nm

Answer

To solve the problem step by step, we need to determine the wavelength of the third line of the series for the `Li^(+2)` ion based on the given conditions.

### Step 1: Determine the values of n1 and n2
We are given two equations:
1. \( n_1 + n_2 = 4 \) (Equation 1)
2. \( n_2 - n_1 = 2 \) (Equation 2)

To find \( n_1 \) and \( n_2 \), we can add both equations:

\[ (n_1 + n_2) + (n_2 - n_1) = 4 + 2 \]
\[ 2n_2 = 6 \quad \Rightarrow \quad n_2 = 3 \]

Now, substitute \( n_2 = 3 \) back into Equation 1:

\[ n_1 + 3 = 4 \quad \Rightarrow \quad n_1 = 1 \]

### Step 2: Identify the transitions
From the values we found:
- \( n_1 = 1 \)
- \( n_2 = 3 \)

The transition is from \( n_2 \) to \( n_1 \), which means the transition is from \( n = 3 \) to \( n = 1 \).

### Step 3: Determine the third line of the series
The problem states we need the third line of the series. In the Lyman series, the transitions are:
- First line: \( n = 2 \to n = 1 \)
- Second line: \( n = 3 \to n = 1 \)
- Third line: \( n = 4 \to n = 1 \)

Thus, for the third line, \( n_2 = 4 \).

### Step 4: Use the Rydberg formula
The Rydberg formula for the wavelength of emitted light is given by:

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

Where:
- \( R \) is the Rydberg constant (\( 1.097 \times 10^7 \, \text{m}^{-1} \))
- \( Z \) is the atomic number of lithium, which is 3.

Substituting \( n_1 = 1 \) and \( n_2 = 4 \):

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \cdot 3^2 \left( \frac{1}{1^2} - \frac{1}{4^2} \right) \]

Calculating \( 3^2 = 9 \):

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \cdot 9 \left( 1 - \frac{1}{16} \right) \]
\[ = 1.097 \times 10^7 \cdot 9 \left( \frac{15}{16} \right) \]
\[ = 1.097 \times 10^7 \cdot \frac{135}{16} \]

Calculating the right-hand side:

\[ \frac{1}{\lambda} = \frac{1.097 \times 10^7 \cdot 135}{16} \]
Calculating \( \frac{1.097 \times 135}{16} \):

\[ = \frac{148.095}{16} \approx 9.2559375 \times 10^6 \, \text{m}^{-1} \]

### Step 5: Calculate the wavelength
Now, take the reciprocal to find \( \lambda \):

\[ \lambda = \frac{1}{9.2559375 \times 10^6} \approx 1.080 \times 10^{-7} \, \text{m} = 10.80 \, \text{nm} \]

### Final Answer
The wavelength of the third line of this series in the `Li^(+2)` ion is approximately \( 10.80 \, \text{nm} \).

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