A gas of hydrogen like ions is prepared in a particular excited state A. if emit photons having wavelength equal to the wavelength of the first line of the lyman series together with photons of five other wavelength identify the gas and find the principal quantum number of the state `A`

Explanation

To solve the problem, we need to identify the gas of hydrogen-like ions and find the principal quantum number of the excited state A. Let's break it down step by step.

### Step 1: Understand the Lyman Series
The Lyman series corresponds to transitions where electrons fall to the n=1 level from higher energy levels (n=2, 3, 4, ...). The first line of the Lyman series corresponds to the transition from n=2 to n=1.

### Step 2: Identify the Total Number of Transitions
The problem states that the gas emits photons with wavelengths equal to the first line of the Lyman series and five other wavelengths. This means there are a total of 6 transitions happening.

### Step 3: Use the Formula for Total Transitions
The total number of transitions (T) that can occur from an excited state n can be calculated using the formula:
\[ T = \frac{n(n-1)}{2} \]
where n is the principal quantum number of the excited state.

### Step 4: Set Up the Equation
Since we know that T = 6 (1 for the Lyman series + 5 other transitions), we can set up the equation:
\[ \frac{n(n-1)}{2} = 6 \]

### Step 5: Solve for n
Multiplying both sides by 2 gives:
\[ n(n-1) = 12 \]
Now, we can rearrange this into a quadratic equation:
\[ n^2 - n - 12 = 0 \]

### Step 6: Factor the Quadratic Equation
We can factor this equation:
\[ (n - 4)(n + 3) = 0 \]
This gives us two possible solutions for n:
1. \( n = 4 \)
2. \( n = -3 \) (not a valid solution since n must be a positive integer)

Thus, the only valid solution is:
\[ n = 4 \]

### Step 7: Identify the Gas
Since the gas is a hydrogen-like ion, we can identify it based on the principal quantum number. Hydrogen-like ions have one electron and can be represented as \( Z \) where \( Z \) is the atomic number. The only hydrogen-like ion that can have an excited state of n=4 is \( He^+ \) (Helium ion).

### Final Answer
The gas is \( He^+ \) (Helium ion), and the principal quantum number of the excited state A is \( n = 4 \).

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