The Bohr of second energy level of `He^(o+)` ion is ……..nm.

Explanation

To find the radius of the second energy level of the `He^(o+)` ion, we can use the formula for the radius of an electron orbit in a hydrogen-like atom:

### Step-by-Step Solution:

1. **Identify the Formula**: The formula for the radius of the nth energy level in a hydrogen-like atom is given by:
\[ r_n = 0.529 \times \frac{n^2}{Z} \text{ angstroms} \]
where \( n \) is the principal quantum number and \( Z \) is the atomic number.

2. **Determine the Values**:
- For the `He^(o+)` ion (which is a helium ion with a single electron), the atomic number \( Z = 2 \).
- We are looking for the second energy level, so \( n = 2 \).

3. **Substitute the Values into the Formula**:
\[ r_2 = 0.529 \times \frac{2^2}{2} \]
Simplifying this:
\[ r_2 = 0.529 \times \frac{4}{2} = 0.529 \times 2 = 1.058 \text{ angstroms} \]

4. **Convert Angstroms to Nanometers**:
- We know that \( 1 \text{ angstrom} = 10^{-1} \text{ nanometers} \).
- Therefore, to convert angstroms to nanometers:
\[ r_2 = 1.058 \text{ angstroms} = 1.058 \times 10^{-1} \text{ nanometers} = 0.1058 \text{ nanometers} \]

5. **Final Answer**:
- Rounding to three significant figures, the radius of the second energy level of the `He^(o+)` ion is approximately:
\[ \text{Answer} = 0.106 \text{ nm} \]