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

Explanation

To find the time period of revolution of the electron in the second excited state of a \( \text{He}^+ \) ion, we can use the relationship derived from Bohr's model of the atom.

### Step-by-step Solution:

1. **Understand the given data**:
- Time period of electron in ground state of hydrogen (\( T_1 \)):
\[ T_1 = 1.60 \times 10^{-16} \text{ s} \]
- For hydrogen, the principal quantum number for the ground state (\( n_1 \)) is 1, and the atomic number (\( z_1 \)) is 1.

2. **Identify the parameters for \( \text{He}^+ \)**:
- For the second excited state of \( \text{He}^+ \), the principal quantum number (\( n_2 \)) is 3 (since the ground state is \( n=1 \), first excited state is \( n=2 \), and second excited state is \( n=3 \)).
- The atomic number (\( z_2 \)) for \( \text{He}^+ \) is 2.

3. **Use the formula for the time period**:
The time period \( T \) is directly proportional to \( \frac{n^3}{z^2} \):
\[ T \propto \frac{n^3}{z^2} \]
Therefore, we can write:
\[ \frac{T_2}{T_1} = \frac{n_2^3}{n_1^3} \cdot \frac{z_1^2}{z_2^2} \]

4. **Substituting the known values**:
\[ \frac{T_2}{T_1} = \frac{3^3}{1^3} \cdot \frac{1^2}{2^2} = \frac{27}{1} \cdot \frac{1}{4} = \frac{27}{4} \]

5. **Calculating \( T_2 \)**:
\[ T_2 = T_1 \cdot \frac{27}{4} = 1.60 \times 10^{-16} \cdot \frac{27}{4} \]
\[ T_2 = 1.60 \times 10^{-16} \cdot 6.75 = 10.8 \times 10^{-16} \text{ s} \]

6. **Final result**:
\[ T_2 = 1.08 \times 10^{-15} \text{ s} \]

### Conclusion:
The time period of revolution of the electron in the second excited state of the \( \text{He}^+ \) ion is \( 1.08 \times 10^{-15} \text{ s} \).