The charge on a particle `Y` is double the charge on particle `X`. These two particles `X` and `Y` after being accelerated through the same potential difference enter a region of uniform magnetic field and describe circular paths of radii `R_(1)` and `R_(2)` respectively. The ratio of the mass of `X` to that of `Y` is

Options

A. `((2R_(1))/(R_(2)))^(2)`
B. `((R_(1))/(2R_(2)))^(2)`
C. `R_(1)^(2)/(2R_(2)^(2)`
D. `(2R_(1))/(R_(2))`

Explanation

To solve the problem, we need to analyze the relationship between the charges, masses, and radii of the circular paths of the two particles \(X\) and \(Y\) when they are subjected to the same potential difference in a magnetic field.

### Step-by-Step Solution:

1. **Understanding the Charges**:
- Let the charge on particle \(X\) be \(q_X\).
- The charge on particle \(Y\) is double that of \(X\), so \(q_Y = 2q_X\).

2. **Kinetic Energy from Potential Difference**:
- When the particles are accelerated through the same potential difference \(V\), the kinetic energy (KE) gained by each particle can be expressed as:
\[ KE_X = q_X \cdot V \]
\[ KE_Y = q_Y \cdot V = 2q_X \cdot V \]

3. **Relating Kinetic Energy to Mass and Velocity**:
- The kinetic energy can also be expressed in terms of mass and velocity:
\[ KE = \frac{1}{2} mv^2 \]
- For particle \(X\):
\[ \frac{1}{2} m_X v_X^2 = q_X \cdot V \quad \Rightarrow \quad v_X^2 = \frac{2q_X \cdot V}{m_X} \]
- For particle \(Y\):
\[ \frac{1}{2} m_Y v_Y^2 = 2q_X \cdot V \quad \Rightarrow \quad v_Y^2 = \frac{4q_X \cdot V}{m_Y} \]

4. **Radius of Circular Motion in a Magnetic Field**:
- The radius \(R\) of the circular path in a magnetic field is given by:
\[ R = \frac{mv}{qB} \]
- For particle \(X\):
\[ R_1 = \frac{m_X v_X}{q_X B} \]
- For particle \(Y\):
\[ R_2 = \frac{m_Y v_Y}{q_Y B} = \frac{m_Y v_Y}{2q_X B} \]

5. **Substituting Velocity Expressions**:
- Substitute \(v_X\) and \(v_Y\) into the radius equations:
\[ R_1 = \frac{m_X \sqrt{\frac{2q_X V}{m_X}}}{q_X B} = \frac{\sqrt{2m_X q_X V}}{q_X B} \]
\[ R_2 = \frac{m_Y \sqrt{\frac{4q_X V}{m_Y}}}{2q_X B} = \frac{\sqrt{4m_Y q_X V}}{2q_X B} \]

6. **Finding the Ratio of Radii**:
- Now, we can find the ratio of the radii:
\[ \frac{R_1}{R_2} = \frac{\sqrt{2m_X q_X V}}{\frac{\sqrt{4m_Y q_X V}}{2}} = \frac{2\sqrt{2m_X}}{\sqrt{4m_Y}} = \frac{2\sqrt{2m_X}}{2\sqrt{m_Y}} = \frac{\sqrt{2m_X}}{\sqrt{m_Y}} \]
- Squaring both sides gives:
\[ \left(\frac{R_1}{R_2}\right)^2 = \frac{2m_X}{m_Y} \]

7. **Finding the Mass Ratio**:
- Rearranging gives:
\[ \frac{m_X}{m_Y} = \frac{1}{2} \left(\frac{R_1}{R_2}\right)^2 \]

### Final Answer:
The ratio of the mass of particle \(X\) to that of particle \(Y\) is:
\[ \frac{m_X}{m_Y} = \frac{1}{2} \left(\frac{R_1}{R_2}\right)^2 \]