The radius of the Bohr orbit in the ground state of hydrogen atom is 0.5 Å. The radius of the 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

Answer

To find the radius of the orbit of the electron in the third excited state of helium, we can use the formula for the radius of the Bohr orbit:

\[ r_n = A_0 \cdot \frac{n^2}{Z} \]

Where:
- \( r_n \) is the radius of the orbit for the principal quantum number \( n \),
- \( A_0 \) is the Bohr radius (given as 0.5 Å),
- \( n \) is the principal quantum number,
- \( Z \) is the atomic number of the atom.

### Step-by-Step Solution:

1. **Identify the values needed for the formula**:
- The Bohr radius \( A_0 \) is given as 0.5 Å.
- The atomic number \( Z \) for helium (He) is 2.
- The third excited state corresponds to \( n = 4 \) (since the ground state is \( n = 1 \), the first excited state is \( n = 2 \), the second excited state is \( n = 3 \), and the third excited state is \( n = 4 \)).

2. **Substitute the values into the formula**:
\[ r_n = A_0 \cdot \frac{n^2}{Z} \]
Substitute \( A_0 = 0.5 \, \text{Å} \), \( n = 4 \), and \( Z = 2 \):
\[ r_4 = 0.5 \cdot \frac{4^2}{2} \]

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

4. **Substitute \( n^2 \) back into the equation**:
\[ r_4 = 0.5 \cdot \frac{16}{2} \]

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

6. **Multiply by the Bohr radius**:
\[ r_4 = 0.5 \cdot 8 = 4 \, \text{Å} \]

### Final Answer:
The radius of the orbit of the electron in the third excited state of helium is **4 Å**.