The radius of the Bohr orbit in the ground state of hydrogen atom is `0.5 Å`. The radius o fthe orbit of the electron in the third excited state of `He^(+)` will be

Options

A. `8 Å`
B. `4 Å`
C. `0.5 Å`
D. `0.25 Å`

Explanation

To find the radius of the orbit of the electron in the third excited state of the helium ion \( He^+ \), we can use the formula for the radius of the nth orbit in a hydrogen-like atom, which is given by:

\[ R_n = R_0 \cdot \frac{n^2}{Z} \]

where:
- \( R_n \) is the radius of the nth orbit,
- \( R_0 \) is the radius of the first orbit in hydrogen (given as \( 0.5 \, \text{Å} \)),
- \( n \) is the principal quantum number (for the third excited state, \( n = 4 \)),
- \( Z \) is the atomic number of the element (for helium, \( Z = 2 \)).

### Step-by-Step Solution:

1. **Identify the values:**
- \( R_0 = 0.5 \, \text{Å} \)
- \( n = 4 \) (for the third excited state)
- \( Z = 2 \) (for helium)

2. **Substitute the values into the formula:**
\[ R_4 = R_0 \cdot \frac{n^2}{Z} \]
\[ R_4 = 0.5 \cdot \frac{4^2}{2} \]

3. **Calculate \( n^2 \):**
\[ n^2 = 4^2 = 16 \]

4. **Substitute \( n^2 \) into the equation:**
\[ R_4 = 0.5 \cdot \frac{16}{2} \]

5. **Simplify the fraction:**
\[ \frac{16}{2} = 8 \]

6. **Calculate \( R_4 \):**
\[ R_4 = 0.5 \cdot 8 = 4 \, \text{Å} \]

Thus, the radius of the orbit of the electron in the third excited state of \( He^+ \) is \( 4 \, \text{Å} \).

### Final Answer:
The radius of the orbit of the electron in the third excited state of \( He^+ \) is \( 4 \, \text{Å} \).