The radius of the` 2^(nd) `orbit of `Li^(2+)` is x. The expected radius of the 3rd orbit of `Be^(3+)` is :

Options

A. 16/27X
B. 4/9X
C. 9/4X
D. 27/16X

Answer

To solve the problem, we need to find the expected radius of the third orbit of \( \text{Be}^{3+} \) given that the radius of the second orbit of \( \text{Li}^{2+} \) is \( x \).

### Step-by-Step Solution:

1. **Understanding the Formula for Radius:**
The radius \( r_n \) of the nth orbit of a hydrogen-like atom is given by the formula:
\[ r_n = \frac{0.529 \, \text{Å} \cdot n^2}{Z} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

2. **Finding the Radius of the 2nd Orbit of \( \text{Li}^{2+} \):**
For \( \text{Li}^{2+} \):
- Atomic number \( Z = 3 \)
- Principal quantum number \( n = 2 \)
Using the formula:
\[ r_2 = \frac{0.529 \cdot 2^2}{3} = \frac{0.529 \cdot 4}{3} = \frac{2.116}{3} = \frac{2.116}{3} \text{ Å} \]
We are given that this radius is equal to \( x \):
\[ x = \frac{2.116}{3} \text{ Å} \]

3. **Finding the Radius of the 3rd Orbit of \( \text{Be}^{3+} \):**
For \( \text{Be}^{3+} \):
- Atomic number \( Z = 4 \)
- Principal quantum number \( n = 3 \)
Using the formula:
\[ r_3 = \frac{0.529 \cdot 3^2}{4} = \frac{0.529 \cdot 9}{4} = \frac{4.761}{4} = 1.19025 \text{ Å} \]

4. **Relating the Radius of \( \text{Be}^{3+} \) to \( x \):**
We know that:
\[ r_2 = \frac{0.529 \cdot 4}{3} = x \]
Now, we can express \( r_3 \) in terms of \( x \):
\[ r_3 = \frac{0.529 \cdot 9}{4} = \frac{3}{4} \cdot \frac{2.116}{3} = \frac{2.116 \cdot 9}{4 \cdot 3} \]
Simplifying:
\[ r_3 = \frac{9}{4} \cdot \frac{x}{3} = \frac{27}{16} x \]

5. **Final Result:**
The expected radius of the 3rd orbit of \( \text{Be}^{3+} \) is:
\[ r_3 = \frac{27}{16} x \]

### Conclusion:
The expected radius of the 3rd orbit of \( \text{Be}^{3+} \) is \( \frac{27}{16} x \).