There are two samples of `H` and `He^+` atom. Both are in some excited state. In hydrogen atom, total number of 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 :

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Balmer Series in Hydrogen
The Balmer series corresponds to transitions of electrons in a hydrogen atom from higher energy levels (n ≥ 3) to the second energy level (n = 2). The number of lines observed in the Balmer series is given as 4. This means there are 4 transitions possible from n = 3, 4, 5, and 6 to n = 2.

### Step 2: Identify the Maximum Energy Transition in Hydrogen
The maximum energy transition in the Balmer series occurs when the electron transitions from the highest energy level (n = 6) to n = 2. The energy of a photon emitted during this transition can be calculated using the formula:
\[ E = -13.6 \, \text{eV} \times Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For hydrogen (Z = 1), this becomes:
\[ E = -13.6 \, \text{eV} \left( \frac{1}{2^2} - \frac{1}{6^2} \right) \]
Calculating this gives:
\[ E = -13.6 \, \text{eV} \left( \frac{1}{4} - \frac{1}{36} \right) \]
\[ = -13.6 \, \text{eV} \left( \frac{9 - 1}{36} \right) \]
\[ = -13.6 \, \text{eV} \left( \frac{8}{36} \right) \]
\[ = -13.6 \, \text{eV} \left( \frac{2}{9} \right) \]
\[ = -3.02 \, \text{eV} \]

### Step 3: Understand the Paschen Series in Helium Ion (He+)
The Paschen series corresponds to transitions from higher energy levels (n ≥ 4) to the third energy level (n = 3). Given that only 1 line is observed in the Paschen series, this indicates that the electron in the He+ ion is transitioning from n = 4 to n = 3.

### Step 4: Calculate the Energy Required to Excite He+
To find the maximum excitation level of the He+ sample, we will use the energy of the photon emitted from the hydrogen atom. The energy required to excite the electron in He+ from n = 4 to a higher level n2 can be calculated using the same energy formula:
\[ E = -13.6 \, \text{eV} \times Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For He+ (Z = 2), we have:
\[ E = -13.6 \, \text{eV} \times 2^2 \left( \frac{1}{4^2} - \frac{1}{n_2^2} \right) \]
Setting this equal to the energy from the hydrogen transition:
\[ 3.02 \, \text{eV} = 13.6 \times 4 \left( \frac{1}{16} - \frac{1}{n_2^2} \right) \]
\[ 3.02 = 54.4 \left( \frac{1}{16} - \frac{1}{n_2^2} \right) \]
Dividing both sides by 54.4:
\[ \frac{3.02}{54.4} = \frac{1}{16} - \frac{1}{n_2^2} \]
Calculating the left side:
\[ 0.0555 = \frac{1}{16} - \frac{1}{n_2^2} \]
\[ 0.0555 = 0.0625 - \frac{1}{n_2^2} \]
Rearranging gives:
\[ \frac{1}{n_2^2} = 0.0625 - 0.0555 = 0.007 \]
Taking the reciprocal:
\[ n_2^2 = \frac{1}{0.007} \approx 142.86 \]
Taking the square root:
\[ n_2 \approx 12 \]

### Conclusion
The maximum excitation level of the He+ sample will be \( n = 12 \).

---