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 find the wavelength of the third line of the series in the `Li^(+2)` ion, we can follow these steps:

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

To solve these equations, we can add them together:

\[ (n_1 + n_2) + (n_2 - n_1) = 4 + 2 \]

This simplifies to:

\[ 2n_2 = 6 \]

Dividing both sides by 2 gives:

\[ n_2 = 3 \]

Now, we can substitute \( n_2 \) back into the first equation to find \( n_1 \):

\[ n_1 + 3 = 4 \]

Thus:

\[ n_1 = 1 \]

### Step 2: Identify the transition for the third line of the series
The problem states that we need to find the wavelength of the third line of the series. In the Lyman series, \( n_1 \) corresponds to 1, and the values of \( n_2 \) can be 2, 3, 4, etc.

The third line of the Lyman series corresponds to the transition from \( n_2 = 4 \) to \( n_1 = 1 \).

### Step 3: Use the Rydberg formula to calculate the wavelength
The Rydberg formula for hydrogen-like ions is given by:

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

Where:
- \( R \) is the Rydberg constant, approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)
- \( Z \) is the atomic number (for \( Li^{+2} \), \( Z = 3 \))
- \( n_1 = 1 \)
- \( n_2 = 4 \)

Substituting the values into the formula:

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

Calculating \( 3^2 = 9 \):

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

Calculating \( 1 - \frac{1}{16} = \frac{15}{16} \):

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times 9 \times \frac{15}{16} \]

Calculating this gives:

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times 9 \times 0.9375 \approx 9.255 \times 10^7 \, \text{m}^{-1} \]

### Step 4: Calculate the wavelength \( \lambda \)
Now, we can find \( \lambda \):

\[ \lambda = \frac{1}{9.255 \times 10^7} \approx 1.080 \times 10^{-8} \, \text{m} \]

To convert this to nanometers (1 nm = \( 10^{-9} \, \text{m} \)):

\[ \lambda \approx 10.80 \, \text{nm} \]

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

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