Consider one `He^+` ion in excited state `(n = 5)`. Which of the following of the following observations will hold true as per the Bohr's model.

Options

A. `10` emission spectral lines will be seen.
B. The ionization energy needed is less than `2 eV`.
C. The longest emitted wavelength is less than `10//R` (`R =` Rydberg's constant).
D. The electronic separtion from the centre of nucleus is more than `6 Å`.

Explanation

To solve the problem regarding the `He^+` ion in an excited state (n = 5) and to determine which observations hold true according to Bohr's model, we will analyze each option step by step.

### Step 1: Calculate the number of emission spectral lines
According to the formula for the number of spectral lines emitted from an excited state:
\[ \text{Number of spectral lines} = \frac{n(n-1)}{2} \]
For `n = 5`:
\[ \text{Number of spectral lines} = \frac{5(5-1)}{2} = \frac{5 \times 4}{2} = 10 \]
**Conclusion**: Option A is correct.

### Step 2: Calculate the ionization energy
The ionization energy can be calculated using the formula:
\[ E_n = -\frac{13.6 \, \text{eV} \times Z^2}{n^2} \]
For `He^+`, \( Z = 2 \) and \( n = 5 \):
\[ E_5 = -\frac{13.6 \times 2^2}{5^2} = -\frac{13.6 \times 4}{25} = -2.176 \, \text{eV} \]
The ionization energy (enthalpy) is the energy required to remove the electron from the ion:
\[ \text{Ionization Energy} = 0 - E_5 = 0 - (-2.176) = 2.176 \, \text{eV} \]
**Conclusion**: The ionization energy is greater than 2 eV, so Option B is incorrect.

### Step 3: Calculate the longest emitted wavelength
Using the Rydberg formula:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For the longest wavelength, we take \( n_1 = 4 \) and \( n_2 = 5 \):
\[ \frac{1}{\lambda} = R \cdot 2^2 \left( \frac{1}{4^2} - \frac{1}{5^2} \right) = 4R \left( \frac{1}{16} - \frac{1}{25} \right) \]
Calculating the fractions:
\[ \frac{1}{16} - \frac{1}{25} = \frac{25 - 16}{400} = \frac{9}{400} \]
Thus:
\[ \frac{1}{\lambda} = 4R \cdot \frac{9}{400} = \frac{36R}{400} = \frac{9R}{100} \]
So, the wavelength \( \lambda \) is:
\[ \lambda = \frac{100}{9R} \]
Since \( \frac{100}{9R} \) is greater than \( \frac{10}{R} \), Option C is incorrect.

### Step 4: Calculate the electronic separation from the nucleus
The radius of the orbit can be calculated using:
\[ R_n = 0.53 \frac{n^2}{Z} \, \text{Å} \]
For `n = 5` and `Z = 2`:
\[ R_5 = 0.53 \frac{5^2}{2} = 0.53 \frac{25}{2} = 0.53 \times 12.5 = 6.625 \, \text{Å} \]
**Conclusion**: The electronic separation from the center of the nucleus is greater than 6 Å, so Option D is correct.

### Final Summary
- Option A: Correct (10 emission spectral lines)
- Option B: Incorrect (Ionization energy is greater than 2 eV)
- Option C: Incorrect (Longest emitted wavelength is greater than \( \frac{10}{R} \))
- Option D: Correct (Electronic separation is more than 6 Å)