What is likely to be principal quantum number for a circular orbit of diameter `20nm` of the hydrogen atom if we assume Bohr orbit be the same as that represented by the principal quantum number?

Options

A. `10`
B. `14`
C. `12`
D. `16`

Explanation

To find the principal quantum number for a circular orbit of diameter 20 nm in a hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Radius**:
The diameter of the orbit is given as 20 nm. Therefore, the radius (R) can be calculated as:
\[ R = \frac{\text{Diameter}}{2} = \frac{20 \text{ nm}}{2} = 10 \text{ nm} \]

2. **Use the Bohr Model Formula**:
According to the Bohr model, the radius of the nth orbit for a hydrogen atom (where Z = 1) is given by the formula:
\[ R_n = 0.0529 \times n^2 \text{ nm} \]
Here, \( R_n \) is the radius in nanometers and \( n \) is the principal quantum number.

3. **Set the Radius Equal to 10 nm**:
We know from step 1 that the radius is 10 nm. Therefore, we can set up the equation:
\[ 10 \text{ nm} = 0.0529 \times n^2 \text{ nm} \]

4. **Solve for n²**:
To find \( n^2 \), we rearrange the equation:
\[ n^2 = \frac{10 \text{ nm}}{0.0529 \text{ nm}} \approx 189.03 \]

5. **Calculate n**:
Now, take the square root of both sides to find \( n \):
\[ n = \sqrt{189.03} \approx 13.75 \]
Since \( n \) must be a whole number, we round it to the nearest whole number:
\[ n \approx 14 \]

### Final Answer:
The principal quantum number \( n \) for the circular orbit of diameter 20 nm in a hydrogen atom is approximately **14**.

---