The time period of revolution of electron in its ground state orbit in a hydrogen atom is `1.60xx 10 ^(-16)s.` The time period of revolution of the electron in its second excited state in a `He ^(+)` ion is:

Options

A. `1.08xx10 ^(-15)s`
B. `1.30 xx10^(-15)s`
C. `2.30 xx10^(-14)s`
D. `2.08xx 10^(-14)s`

Explanation

To solve the problem, we need to find the time period of revolution of the electron in the second excited state of a He⁺ ion, given the time period of the electron in the ground state of a hydrogen atom.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Time period of electron in ground state of hydrogen (T₁):
\[ T_1 = 1.6 \times 10^{-16} \text{ s} \]
- For hydrogen (H):
- Principal quantum number (n₁) = 1 (ground state)
- Atomic number (Z₁) = 1
- For He⁺ ion:
- Principal quantum number (n₂) = 3 (second excited state)
- Atomic number (Z₂) = 2

2. **Use the Relationship Between Time Periods:**
The time period (T) of an electron in an orbit is directly proportional to \( \frac{n^3}{Z^2} \):
\[ T \propto \frac{n^3}{Z^2} \]
Therefore, we can write the ratio of the time periods for hydrogen and He⁺ as:
\[ \frac{T_1}{T_2} = \frac{n_1^3 / Z_1^2}{n_2^3 / Z_2^2} \]

3. **Substitute the Known Values:**
Substituting the values we have:
\[ \frac{T_1}{T_2} = \frac{1^3 / 1^2}{3^3 / 2^2} \]
This simplifies to:
\[ \frac{T_1}{T_2} = \frac{1 / 1}{27 / 4} = \frac{4}{27} \]

4. **Rearranging to Find T₂:**
Rearranging the equation gives:
\[ T_2 = T_1 \cdot \frac{27}{4} \]
Now substituting \( T_1 \):
\[ T_2 = 1.6 \times 10^{-16} \cdot \frac{27}{4} \]

5. **Calculating T₂:**
Performing the multiplication:
\[ T_2 = 1.6 \times 10^{-16} \cdot 6.75 = 1.08 \times 10^{-15} \text{ s} \]

6. **Final Answer:**
The time period of revolution of the electron in its second excited state in a He⁺ ion is:
\[ T_2 = 1.08 \times 10^{-15} \text{ s} \]

### Conclusion:
The correct option is **A: \( 1.08 \times 10^{-15} \text{ s} \)**.