In a hydrogen like ionized atom a single electron is orbiting around the stationary positive charge. If a spectral line of `lambda` equal to `4861Å` is observed due to transition from `n=12` to `n-6`. What is the wavelength of a spectral line due to transition from `n=9` to `n=6` and also identify the element.

Explanation

To solve the problem step by step, we will first analyze the given data and then use the formulas related to energy transitions in hydrogen-like atoms.

### Step 1: Identify the given data
We are given:
- Wavelength of the spectral line for the transition from \( n = 12 \) to \( n = 6 \): \( \lambda_1 = 4861 \, \text{Å} \)
- Initial state for the second transition: \( n_i = 9 \)
- Final state for both transitions: \( n_f = 6 \)

### Step 2: Calculate the energy difference for the first transition
The energy difference for a transition in a hydrogen-like atom can be expressed as:
\[ \Delta E = E_0 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
For the transition from \( n = 12 \) to \( n = 6 \):
\[ \Delta E_1 = E_0 \left( \frac{1}{6^2} - \frac{1}{12^2} \right) \]
Calculating the values:
\[ \Delta E_1 = E_0 \left( \frac{1}{36} - \frac{1}{144} \right) = E_0 \left( \frac{4 - 1}{144} \right) = \frac{3E_0}{144} = \frac{E_0}{48} \]

### Step 3: Relate energy difference to wavelength
Using the relation between energy and wavelength:
\[ \Delta E = \frac{hc}{\lambda} \]
For the first transition:
\[ \Delta E_1 = \frac{hc}{\lambda_1} \]
Substituting \( \lambda_1 = 4861 \, \text{Å} \):
\[ \frac{E_0}{48} = \frac{hc}{4861} \]

### Step 4: Calculate the energy difference for the second transition
For the transition from \( n = 9 \) to \( n = 6 \):
\[ \Delta E_2 = E_0 \left( \frac{1}{6^2} - \frac{1}{9^2} \right) \]
Calculating the values:
\[ \Delta E_2 = E_0 \left( \frac{1}{36} - \frac{1}{81} \right) = E_0 \left( \frac{9 - 4}{144} \right) = \frac{5E_0}{144} \]

### Step 5: Relate the two energy differences
Using the ratio of the energy differences:
\[ \frac{\Delta E_1}{\Delta E_2} = \frac{\frac{E_0}{48}}{\frac{5E_0}{144}} = \frac{144}{240} = \frac{3}{5} \]
Thus, we can express the wavelengths:
\[ \frac{\lambda_2}{\lambda_1} = \frac{\Delta E_1}{\Delta E_2} \implies \lambda_2 = \lambda_1 \cdot \frac{5}{3} \]

### Step 6: Calculate the wavelength for the second transition
Substituting \( \lambda_1 = 4861 \, \text{Å} \):
\[ \lambda_2 = 4861 \cdot \frac{5}{3} = 8101.67 \, \text{Å} \]

### Step 7: Identify the element
From the first transition, we calculated \( E_0 \) and found \( Z \) using:
\[ E_0 = 13.6 \times Z^2 \quad \text{and} \quad E_0 \approx 123 \, \text{eV} \]
Thus:
\[ Z^2 = \frac{123}{13.6} \approx 9 \implies Z = 3 \]
This means the element is Lithium (Li) in its ionized form, \( \text{Li}^{2+} \).

### Final Answer
The wavelength of the spectral line due to the transition from \( n = 9 \) to \( n = 6 \) is approximately \( 8101.67 \, \text{Å} \), and the element is Lithium (\( \text{Li}^{2+} \)).

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