Two particles \( X \) and \( Y \) having equal charges are being accelerated through the same potential difference. Thereafter, they enter normally in a region of uniform magnetic field and describe circular paths of radii \( R_1 \) and \( R_2 \) respectively. The mass ratio of \( X \) and \( Y \) is:
- \( \left( \frac{R_2}{R_1} \right)^2 \)
- \( \left( \frac{R_1}{R_2} \right)^2 \)
- \( \frac{R_1}{R_2} \)
- \( \frac{R_2}{R_1} \)
The Correct Option is B
Approach Solution - 1
To solve the problem, we need to determine the mass ratio of two particles \( X \) and \( Y \) that have equal charges and are accelerated through the same potential difference. They enter a uniform magnetic field, resulting in circular paths with radii \( R_1 \) (for \( X \)) and \( R_2 \) (for \( Y \)).
The formula for the radius \( R \) of the circular path described by a charged particle in a magnetic field is given by:
\(R = \frac{mv}{qB}\)
Where:
- \(m\) is the mass of the particle.
- \(v\) is the velocity of the particle.
- \(q\) is the charge of the particle.
- \(B\) is the magnetic field strength.
The kinetic energy \(\left( KE \right)\) of a particle accelerated through a potential difference \( V \) is:
\(\frac{1}{2}mv^2 = qV\)
From the equation of kinetic energy, we find the velocity:
\(v = \sqrt{\frac{2qV}{m}}\)
Substituting this velocity back into the expression for \(R\):
\(R = \frac{\sqrt{\frac{2qV}{m}} \cdot m}{qB} = \frac{m}{qB} \cdot \sqrt{\frac{2qV}{m}}\)
Simplifying, we get:
\(R = \sqrt{\frac{2Vm}{qB^2}}\)
From this equation, for particles \( X \) and \( Y \), whose radii are \( R_1 \) and \( R_2 \) respectively, and using the fact that \( q, V, \) and \( B \) are the same for both particles, we have:
\(R_1 = \sqrt{\frac{2Vm_1}{qB^2}} \quad \text{and} \quad R_2 = \sqrt{\frac{2Vm_2}{qB^2}}\)
Equating the expressions for \( R_1^2 \) and \( R_2^2 \):
\(R_1^2 = \frac{2Vm_1}{qB^2} \quad \text{and} \quad R_2^2 = \frac{2Vm_2}{qB^2}\)
To find the mass ratio \(\frac{m_1}{m_2}\), we divide these two equations:
\(\frac{R_1^2}{R_2^2} = \frac{m_1}{m_2}\)
Hence, the mass ratio \(\frac{m_1}{m_2}\) is given by:
\(\frac{m_1}{m_2} = \left( \frac{R_1}{R_2} \right)^2\)
Thus, the correct answer is:
\(<\left( \frac{R_1}{R_2} \right)^2 \)
Approach Solution -2
Given:
- The particles have equal charges \( q \).
- They are accelerated through the same potential difference \( V \).
- Radii of circular paths in a magnetic field are \( R_1 \) for particle \( X \) and \( R_2 \) for particle \( Y \).
Step 1. Relate radius to mass and velocity:
For a particle moving in a circular path in a magnetic field, the radius \( R \) is given by:
\(R = \frac{mv}{qB}\)
where:
- \( m \) is the mass of the particle,
- \( v \) is the velocity after acceleration,
- \( q \) is the charge, and
- \( B \) is the magnetic field strength.
Step 2. Express \( v \) in terms of \( V \):
Since each particle is accelerated through the same potential difference \( V \), the kinetic energy gained by each particle is:
\(\frac{1}{2} mv^2 = qV\)
Solving for \( v \), we get:
\(v = \sqrt{\frac{2qV}{m}}\)
Step 3. Substitute \( v \) into the radius formula:
\( R = \frac{m}{qB} \cdot \sqrt{\frac{2qV}{m}} = \frac{\sqrt{2m \cdot qV}}{qB}\)
Step 4. Determine the ratio of radii for particles \( X \) and \( Y \):
\(\frac{R_1}{R_2} = \frac{\sqrt{2m_X \cdot qV}}{\sqrt{2m_Y \cdot qV}} = \sqrt{\frac{m_X}{m_Y}}\)
Step 5. Solve for the mass ratio:
Squaring both sides, we get:
\(\frac{m_X}{m_Y} = \left(\frac{R_1}{R_2}\right)^2\)
Thus, the mass ratio \( \frac{m_X}{m_Y} \) is \( \left(\frac{R_1}{R_2}\right)^2 \).
The Correct Answer is: \( \left( \frac{R_1}{R_2} \right)^2 \)
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