The time period of revolution in the `3^(rd)` orbit of `Li^(2+)` ion is x sec. The time period of revolution in the `2^(nd)` orbit of `He^(+) ion, should be

Options

A. x sec
B. `3/2 x sec`
C. `2/3 x sec`
D. `8/27 x sec`

Explanation

To find the time period of revolution in the second orbit of the He\(^+\) ion, we can use the relationship between the time period, the principal quantum number \(n\), and the atomic number \(Z\). The time period \(T\) is proportional to \(\frac{n^3}{Z^2}\).

### Step-by-Step Solution:

1. **Identify the time period for Li\(^{2+}\)**:
- For the Li\(^{2+}\) ion in the third orbit, we have:
\[ T_{Li^{2+}} \propto \frac{n^3}{Z^2} \]
- Here, \(n = 3\) and \(Z = 3\) (since lithium has an atomic number of 3).
- Therefore:
\[ T_{Li^{2+}} \propto \frac{3^3}{3^2} = \frac{27}{9} = 3 \]
- Given that this time period is \(x\) seconds, we can express it as:
\[ T_{Li^{2+}} = k \cdot 3 \quad \text{(where \(k\) is a constant)} \]
- Thus, we have:
\[ x = k \cdot 3 \]

2. **Identify the time period for He\(^+\)**:
- For the He\(^+\) ion in the second orbit, we have:
\[ T_{He^{+}} \propto \frac{n^3}{Z^2} \]
- Here, \(n = 2\) and \(Z = 2\) (since helium has an atomic number of 2).
- Therefore:
\[ T_{He^{+}} \propto \frac{2^3}{2^2} = \frac{8}{4} = 2 \]
- We can express this as:
\[ T_{He^{+}} = k \cdot 2 \]

3. **Relate the two time periods**:
- Now, we can set up the ratio of the time periods:
\[ \frac{T_{Li^{2+}}}{T_{He^{+}}} = \frac{3}{2} \]
- Substituting the expressions for \(T_{Li^{2+}}\) and \(T_{He^{+}}\):
\[ \frac{x}{T_{He^{+}}} = \frac{3}{2} \]
- Rearranging gives:
\[ T_{He^{+}} = \frac{2}{3}x \]

4. **Final result**:
- Therefore, the time period of revolution in the second orbit of the He\(^+\) ion is:
\[ T_{He^{+}} = \frac{2}{3}x \]

### Conclusion:
The time period of revolution in the second orbit of the He\(^+\) ion is \(\frac{2}{3}x\) seconds.