If radius of `2^(nd)` orbit is x then di-Broglie wavelength in`4^(th)`orbit is given by

Options

A. `8 pi x `
B. `2pi x `
C. `4 pi x`
D. `6 pi x `

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We know that the radius of the second orbit (n=2) is given as \( x \).

2. **Using Bohr's Model for Radius**:
- The formula for the radius of the nth orbit in Bohr's model is given by:
\[ r_n = \frac{0.529 n^2}{Z} \text{ angstroms} \]
- For the second orbit (n=2), we can express this as:
\[ r_2 = \frac{0.529 \times 2^2}{Z} = \frac{0.529 \times 4}{Z} \]
- We know that \( r_2 = x \), so we have:
\[ \frac{0.529 \times 4}{Z} = x \]

3. **Finding the Radius of the Fourth Orbit**:
- Now, we will find the radius for the fourth orbit (n=4):
\[ r_4 = \frac{0.529 \times 4^2}{Z} = \frac{0.529 \times 16}{Z} \]
- We can factor out the 4 from \( r_4 \):
\[ r_4 = 4 \times \frac{0.529 \times 4}{Z} = 4 \times r_2 \]
- Since \( r_2 = x \), we have:
\[ r_4 = 4x \]

4. **Calculating the de Broglie Wavelength**:
- The de Broglie wavelength (\( \lambda \)) is related to the radius and the principal quantum number (n) by the formula:
\[ 2\pi r = n\lambda \]
- For the fourth orbit (n=4), we can write:
\[ 2\pi r_4 = 4\lambda_4 \]
- Rearranging gives:
\[ \lambda_4 = \frac{2\pi r_4}{4} \]
- Substituting \( r_4 = 4x \):
\[ \lambda_4 = \frac{2\pi (4x)}{4} = 2\pi x \]

5. **Final Answer**:
- Therefore, the de Broglie wavelength in the fourth orbit is:
\[ \lambda_4 = 2\pi x \]

### Conclusion:
The correct option is **B) \( 2\pi x \)**.