Difference in radius of 3rd and 4th orbit in `He^+` ion in `(R_0)_1` and in `Li^(2+)` ion is `(R_0)_2` then ratio of `(R_0)_1` TO `(R_0)_2` is

Options

A. 1:2
B. 2:1
C. 3:2
D. 2:3

Answer

To solve the problem, we need to find the difference in the radius of the 3rd and 4th orbits for the `He^+` and `Li^(2+)` ions, and then calculate the ratio of these differences.

### Step-by-Step Solution:

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

2. **Calculate for `He^+` (Z = 2)**:
- For the 4th orbit (\( n = 4 \)):
\[ R_4 = \frac{0.529 \times 4^2}{2} = \frac{0.529 \times 16}{2} = \frac{8.464}{2} = 4.232 \, \text{Å} \]
- For the 3rd orbit (\( n = 3 \)):
\[ R_3 = \frac{0.529 \times 3^2}{2} = \frac{0.529 \times 9}{2} = \frac{4.761}{2} = 2.3805 \, \text{Å} \]
- Difference in radius (\( R_4 - R_3 \)):
\[ R_1 = R_4 - R_3 = 4.232 - 2.3805 = 1.8515 \, \text{Å} \]

3. **Calculate for `Li^(2+)` (Z = 3)**:
- For the 4th orbit (\( n = 4 \)):
\[ R_4 = \frac{0.529 \times 4^2}{3} = \frac{0.529 \times 16}{3} = \frac{8.464}{3} = 2.8213 \, \text{Å} \]
- For the 3rd orbit (\( n = 3 \)):
\[ R_3 = \frac{0.529 \times 3^2}{3} = \frac{0.529 \times 9}{3} = \frac{4.761}{3} = 1.587 \, \text{Å} \]
- Difference in radius (\( R_4 - R_3 \)):
\[ R_2 = R_4 - R_3 = 2.8213 - 1.587 = 1.2343 \, \text{Å} \]

4. **Calculate the Ratio**:
- Now we find the ratio \( \frac{R_1}{R_2} \):
\[ \frac{R_1}{R_2} = \frac{1.8515}{1.2343} \]
- To simplify, we can approximate:
\[ \frac{R_1}{R_2} \approx \frac{1.8515}{1.2343} \approx 1.5 \approx \frac{3}{2} \]

5. **Final Answer**:
The ratio of \( R_0 \) for `He^+` to `Li^(2+)` is:
\[ \frac{R_0}_1 : \frac{R_0}_2 = 3 : 2 \]