A hydrogne like atom is excited using a radiation . Consequently, six spectral line are observed in the spectrum. The wavelegth of emission radiation is found to be equal or smaller than the radiation used for excitation. This concludes that the gas was initially at

Options

A. Ground state
B. First excited state
C. Second excited state
D. Third excited state

Explanation

To solve the problem, we need to analyze the information given about the hydrogen-like atom and the spectral lines observed. Let's break it down step by step.

### Step 1: Understand the number of spectral lines
The problem states that six spectral lines are observed when the hydrogen-like atom is excited. The number of spectral lines (N) that can be observed from transitions between energy levels is given by the formula:

\[ N = \frac{n(n-1)}{2} \]

where \( n \) is the number of energy levels involved in the transitions.

### Step 2: Set up the equation
Given that \( N = 6 \), we can set up the equation:

\[ 6 = \frac{n(n-1)}{2} \]

### Step 3: Solve for \( n \)
To find \( n \), we can rearrange the equation:

\[ n(n-1) = 12 \]

Now we can solve for \( n \) by testing integer values:

- For \( n = 4 \):
\[ 4(4-1) = 4 \times 3 = 12 \] (This works)

Thus, \( n = 4 \).

### Step 4: Determine the final state of the atom
In a hydrogen-like atom, the transitions can occur from the initial state to the final states. The maximum principal quantum number \( n \) that can be reached is 4, which means the atom can be in states \( n = 1, 2, 3, \) or \( 4 \).

### Step 5: Identify the initial state
Since the atom can emit radiation with wavelengths equal to or smaller than the excitation radiation, we need to determine the initial state of the atom. The transitions that can occur from the excited state \( n = 4 \) to lower states (1, 2, 3) will produce spectral lines.

The initial state of the atom must be the second excited state, which corresponds to \( n = 3 \) (since the ground state is \( n = 1 \), the first excited state is \( n = 2 \), and the second excited state is \( n = 3 \)).

### Conclusion
Thus, we conclude that the gas was initially at the **second excited state**.

### Final Answer
The gas was initially at the **second excited state (n = 3)**.

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