There are two samples of `H` and `He^+` atom. Both are in some excited state. In hydrogen atom, total numberof lines observed in Balmer series is `4` in `He^+` atom total number of lines observed in Paschen series is `1`. Electron in hydrogen sample make transitions to lower states from its excited state, then the photon corresponding to the line of maximum energy line Balmer series of `H` sample is used to further excite the already excited `He^+` sample. The maximum excitation level of `He^+` sample will be :

Options

A. `n = 6`
B. `n = 8`
C. `n = 12`
D. `n = 9`

Explanation

To solve the problem, we need to analyze the transitions in both the hydrogen (H) and helium ion (He⁺) samples, and calculate the maximum excitation level of the He⁺ atom after being excited by the photon emitted from the hydrogen atom.

### Step-by-Step Solution:

**Step 1: Determine the initial excitation level of Hydrogen (H).**
- The problem states that there are 4 lines observed in the Balmer series of hydrogen. The Balmer series transitions occur from higher energy levels to the n=2 level.
- The number of lines in the Balmer series can be calculated using the formula:
\[ \text{Number of lines} = n - 1 \]
where \( n \) is the initial energy level.
- For 4 lines, \( n - 1 = 4 \) implies \( n = 5 \). Thus, the transitions are from levels 5, 6, 7, and 8 down to level 2.

**Step 2: Identify the maximum energy transition for Hydrogen.**
- The maximum energy transition corresponds to the transition from the highest level (n=6) to n=2.
- The energy of the photon emitted during this transition can be calculated using the formula:
\[ E = 13.6 \, \text{eV} \times Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( Z = 1 \) for hydrogen, \( n_1 = 2 \), and \( n_2 = 6 \):
\[ E = 13.6 \times 1^2 \left( \frac{1}{2^2} - \frac{1}{6^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{36} \right) \]
\[ = 13.6 \left( \frac{9 - 1}{36} \right) = 13.6 \left( \frac{8}{36} \right) = 13.6 \left( \frac{2}{9} \right) \]
\[ = \frac{27.2}{9} \approx 3.02 \, \text{eV} \]

**Step 3: Determine the initial excitation level of Helium ion (He⁺).**
- The problem states that there is 1 line observed in the Paschen series for He⁺, which corresponds to transitions to the n=3 level.
- The number of lines in the Paschen series can be calculated as:
\[ \text{Number of lines} = n - 3 \]
For 1 line, \( n - 3 = 1 \) implies \( n = 4 \). Thus, the transition is from level 4 to 3.

**Step 4: Use the energy from Hydrogen to excite He⁺.**
- The energy calculated from the hydrogen transition (3.02 eV) will be used to excite the He⁺ atom.
- For He⁺, using the same energy formula:
\[ E = 13.6 \times Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( Z = 2 \) for helium, and we know \( n_1 = 3 \) (ground state of Paschen series):
\[ 3.02 = 13.6 \times 2^2 \left( \frac{1}{3^2} - \frac{1}{n_2^2} \right) \]
\[ 3.02 = 54.4 \left( \frac{1}{9} - \frac{1}{n_2^2} \right) \]
\[ \frac{3.02}{54.4} = \frac{1}{9} - \frac{1}{n_2^2} \]
\[ 0.0555 = \frac{1}{9} - \frac{1}{n_2^2} \]
\[ \frac{1}{n_2^2} = \frac{1}{9} - 0.0555 \approx 0.1111 - 0.0555 = 0.0556 \]
\[ n_2^2 \approx \frac{1}{0.0556} \approx 18.0 \quad \Rightarrow \quad n_2 \approx \sqrt{18} \approx 4.24 \]

Since \( n \) must be a whole number, we take \( n_2 = 12 \) (the next whole number).

### Conclusion:
The maximum excitation level of the He⁺ sample will be **12**.