An electron is revolving in the `n^(th)` orbit of radius `4.2 Å`, then the value of n is `(r1=0.529 Å)`

Options

A. 4
B. 5
C. 6
D. 3

Explanation

To find the value of \( n \) for an electron revolving in the \( n^{th} \) orbit of radius \( 4.2 \, \text{Å} \), we can use the formula that relates the radius of the orbit to the principal quantum number \( n \).

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The radius of the \( n^{th} \) orbit \( R_n \) is given by the formula:
\[ R_n = n^2 \cdot R_1 \]
where \( R_1 \) is the radius of the first orbit (Bohr radius).

2. **Substituting Known Values**:
We know:
- \( R_n = 4.2 \, \text{Å} = 4.2 \times 10^{-10} \, \text{m} \)
- \( R_1 = 0.529 \, \text{Å} = 0.529 \times 10^{-10} \, \text{m} \)

Plugging these values into the equation gives:
\[ 4.2 \times 10^{-10} = n^2 \cdot (0.529 \times 10^{-10}) \]

3. **Rearranging the Equation**:
To find \( n^2 \), we can rearrange the equation:
\[ n^2 = \frac{4.2 \times 10^{-10}}{0.529 \times 10^{-10}} \]

4. **Calculating \( n^2 \)**:
Simplifying the right side:
\[ n^2 = \frac{4.2}{0.529} \]
Calculating this gives:
\[ n^2 \approx 7.94 \]

5. **Finding \( n \)**:
Taking the square root to find \( n \):
\[ n = \sqrt{7.94} \approx 2.81 \]

6. **Rounding to the Nearest Whole Number**:
Since \( n \) must be a whole number, we round \( 2.81 \) to the nearest whole number:
\[ n = 3 \]

### Final Answer:
Thus, the value of \( n \) is \( 3 \).

---