Two particles X and Y having equal charges, 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 masses of X and Y is

Options

A. `(R_1//R_2)^(1//2)`
B. `(R_2//R_1)`
C. `(R_1//R_2)^2`
D. `(R_1//R_2)`

Explanation

To solve the problem, we need to find the ratio of the masses of two particles X and Y, which have equal charges and are accelerated through the same potential difference before entering a magnetic field and moving in circular paths of radii \( R_1 \) and \( R_2 \) respectively.

### Step-by-Step Solution:

1. **Understanding Kinetic Energy**:
When a charged particle is accelerated through a potential difference \( V \), it gains kinetic energy given by:
\[ KE = qV \]
where \( q \) is the charge of the particle.

2. **Relating Kinetic Energy to Velocity**:
The kinetic energy can also be expressed in terms of mass \( m \) and velocity \( v \):
\[ KE = \frac{1}{2} mv^2 \]
Setting these two expressions for kinetic energy equal gives:
\[ \frac{1}{2} mv^2 = qV \]

3. **Solving for Velocity**:
Rearranging the equation to solve for \( v \):
\[ v = \sqrt{\frac{2qV}{m}} \]
This is our **Equation (1)**.

4. **Motion in a Magnetic Field**:
When a charged particle moves in a magnetic field, it experiences a magnetic force that acts as the centripetal force, allowing it to move in a circular path. The centripetal force can be expressed as:
\[ F_c = \frac{mv^2}{r} \]
where \( r \) is the radius of the circular path.

5. **Magnetic Force**:
The magnetic force acting on the particle is given by:
\[ F_m = qvB \]
where \( B \) is the magnetic field strength.

6. **Setting Forces Equal**:
For circular motion, the centripetal force is provided by the magnetic force:
\[ \frac{mv^2}{r} = qvB \]

7. **Solving for Velocity Again**:
Rearranging gives:
\[ v = \frac{qBr}{m} \]
This is our **Equation (2)**.

8. **Equating the Two Expressions for Velocity**:
Now we have two expressions for \( v \):
\[ \sqrt{\frac{2qV}{m}} = \frac{qBr}{m} \]

9. **Squaring Both Sides**:
Squaring both sides to eliminate the square root:
\[ \frac{2qV}{m} = \frac{q^2B^2r^2}{m^2} \]

10. **Rearranging for Mass**:
Rearranging gives:
\[ m = \frac{q^2B^2r^2}{2qV} \]
Simplifying this gives:
\[ m \propto r^2 \]

11. **Finding the Ratio of Masses**:
For particles X and Y, since they have equal charges and are accelerated through the same potential difference:
\[ \frac{m_X}{m_Y} = \frac{R_1^2}{R_2^2} \]

### Final Result:
Thus, the ratio of the masses of particles X and Y is:
\[ \frac{m_X}{m_Y} = \frac{R_1^2}{R_2^2} \]