If lambda is the wavelength of hydrogen atom from the transition `n=3 to n=1`,then what is the wavelength for doubly ionised lithium ion for same transition?

Options

A. `(lambda)/(3)`
B. `3lambda`
C. `(lambda)/(9)`
D. `9lambda`

Answer

To solve the problem, we need to find the wavelength of the transition from \( n = 3 \) to \( n = 1 \) for a doubly ionized lithium ion (\( \text{Li}^{2+} \)) given the wavelength for the same transition in a hydrogen atom.

### Step-by-Step Solution:

1. **Understand the Formula**:
The wavelength (\( \lambda \)) for a transition in a hydrogen-like atom can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number,
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final states respectively.

2. **Identify the Values for Hydrogen**:
For hydrogen (\( Z_H = 1 \)):
- \( n_1 = 1 \)
- \( n_2 = 3 \)
Thus, the wavelength for hydrogen can be expressed as:
\[ \frac{1}{\lambda_H} = R(1^2) \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R \left( 1 - \frac{1}{9} \right) = R \left( \frac{8}{9} \right) \]

3. **Calculate the Wavelength for Lithium Ion**:
For doubly ionized lithium (\( Z_{Li} = 3 \)):
- The same values of \( n_1 \) and \( n_2 \) apply.
Therefore, for lithium:
\[ \frac{1}{\lambda_{Li}} = R(3^2) \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R(9) \left( \frac{8}{9} \right) = R(8) \]

4. **Relate the Wavelengths**:
Now, we can relate the wavelengths of hydrogen and lithium:
\[ \frac{1}{\lambda_H} = R \left( \frac{8}{9} \right) \quad \text{and} \quad \frac{1}{\lambda_{Li}} = R(8) \]
To find the ratio:
\[ \frac{\lambda_{Li}}{\lambda_H} = \frac{R(8)}{R \left( \frac{8}{9} \right)} = \frac{8}{\frac{8}{9}} = 9 \]
Thus, we have:
\[ \lambda_{Li} = \frac{\lambda_H}{9} \]

5. **Final Answer**:
Therefore, the wavelength for the doubly ionized lithium ion for the transition \( n = 3 \) to \( n = 1 \) is:
\[ \lambda_{Li} = \frac{\lambda_H}{9} \]

### Conclusion:
The answer is \( \frac{\lambda}{9} \).