Radius of an electron moving in a circle in constant magnetic field is two times that of an `alpha` particle in the same field. Then de-Broglie wavelength of electrons is x-times of the `alpha`-particle Here x is

Options

A. 2
B. `(1)/(2)`
C. 1
D. 4

Answer

To solve the problem, we need to find the relationship between the de-Broglie wavelengths of an electron and an alpha particle given that the radius of the electron's path in a magnetic field is twice that of the alpha particle's path.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- Let \( r_e \) be the radius of the electron's path.
- Let \( r_\alpha \) be the radius of the alpha particle's path.
- According to the problem, \( r_e = 2 r_\alpha \).

2. **Using the Formula for Radius in a Magnetic Field:**
- The radius \( r \) of a charged particle moving in a magnetic field \( B \) is given by the formula:
\[ r = \frac{p}{Bq} \]
where \( p \) is the momentum, \( B \) is the magnetic field strength, and \( q \) is the charge of the particle.

3. **Applying the Formula for Both Particles:**
- For the electron:
\[ r_e = \frac{p_e}{B e} \]
- For the alpha particle (which has a charge of \( 2e \)):
\[ r_\alpha = \frac{p_\alpha}{B (2e)} \]

4. **Setting Up the Equation:**
- From the relationship \( r_e = 2 r_\alpha \), we can substitute the expressions for \( r_e \) and \( r_\alpha \):
\[ \frac{p_e}{B e} = 2 \left( \frac{p_\alpha}{B (2e)} \right) \]

5. **Simplifying the Equation:**
- Cancel \( B \) and \( e \) from both sides:
\[ p_e = 2 \left( \frac{p_\alpha}{2} \right) \]
- This simplifies to:
\[ p_e = p_\alpha \]

6. **Finding the De-Broglie Wavelengths:**
- The de-Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
- For the electron:
\[ \lambda_e = \frac{h}{p_e} \]
- For the alpha particle:
\[ \lambda_\alpha = \frac{h}{p_\alpha} \]

7. **Taking the Ratio of the Wavelengths:**
- The ratio of the de-Broglie wavelengths is:
\[ \frac{\lambda_e}{\lambda_\alpha} = \frac{h/p_e}{h/p_\alpha} = \frac{p_\alpha}{p_e} \]
- Since we found that \( p_e = p_\alpha \):
\[ \frac{\lambda_e}{\lambda_\alpha} = \frac{p_\alpha}{p_\alpha} = 1 \]

8. **Conclusion:**
- Therefore, the de-Broglie wavelength of the electron is equal to that of the alpha particle:
\[ \lambda_e = 1 \cdot \lambda_\alpha \]
- Thus, \( x = 1 \).

### Final Answer:
The value of \( x \) is \( 1 \).