In `He^(o+)` ion sample `e^(-)` is in ground state. If photon of energy 52.24 eV is given to the sample all the atom goes to higher energy state. It again falls back up to ground state. If it is not emitting any lines in Balmer series then what is the maximum possible number of spectral lines observed

Answer

To solve the problem, we need to determine the maximum number of spectral lines that can be observed when an electron in a He⁺ ion transitions from an excited state back to the ground state. We know that the photon energy provided is 52.24 eV and that the electron is initially in the ground state.

### Step-by-Step Solution:

1. **Identify the Energy Levels of He⁺:**
The energy levels of a hydrogen-like atom (like He⁺) can be calculated using the formula:
\[ E_n = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
where \( Z \) is the atomic number (for He, \( Z = 2 \)) and \( n \) is the principal quantum number.

2. **Calculate the Maximum Excitation Level:**
Given the energy of the photon is 52.24 eV, we set this equal to the energy difference between the ground state (n=1) and the excited state (n=i):
\[ 52.24 = 13.6 \cdot 2^2 \left( \frac{1}{1^2} - \frac{1}{n_i^2} \right) \]
Simplifying this, we have:
\[ 52.24 = 54.4 \left( 1 - \frac{1}{n_i^2} \right) \]
\[ \frac{52.24}{54.4} = 1 - \frac{1}{n_i^2} \]
\[ 0.96 = 1 - \frac{1}{n_i^2} \]
\[ \frac{1}{n_i^2} = 0.04 \quad \Rightarrow \quad n_i^2 = 25 \quad \Rightarrow \quad n_i = 5 \]

3. **Determine Possible Transitions:**
The electron can transition from the excited state (n=5) to the ground state (n=1). The possible transitions from n=5 to lower energy levels are:
- 5 → 4
- 5 → 3
- 5 → 2
- 5 → 1

Additionally, we can also have:
- 4 → 3
- 4 → 2
- 4 → 1
- 3 → 2
- 3 → 1
- 2 → 1

4. **Count the Total Possible Transitions:**
The total number of transitions from n=5 down to n=1 can be calculated using the formula for combinations:
\[ \text{Total lines} = \frac{n(n-1)}{2} \]
where \( n \) is the number of energy levels involved. Here, \( n = 5 \):
\[ \text{Total lines} = \frac{5(5-1)}{2} = \frac{5 \cdot 4}{2} = 10 \]

5. **Account for the Balmer Series:**
The Balmer series involves transitions that end at n=2. Since we are told that no lines are emitted in the Balmer series, we must exclude transitions that end at n=2:
- 5 → 2
- 4 → 2
- 3 → 2

This gives us 3 transitions that we need to exclude from our total.

6. **Calculate the Final Number of Spectral Lines:**
Therefore, the maximum number of spectral lines observed will be:
\[ \text{Observed lines} = \text{Total lines} - \text{Excluded lines} = 10 - 3 = 7 \]

### Final Answer:
The maximum possible number of spectral lines observed is **7**.