Question

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:

• A. $\left( \frac{R_2}{R_1} \right)^2$
• B. $\left( \frac{R_1}{R_2} \right)^2$
• C. $\frac{R_1}{R_2}$
• D. $\frac{R_2}{R_1}$

Answer 1

Answer: (B)

$\left( \frac{R_1}{R_2} \right)^2$

Answer 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: $\left( \frac{R_1}{R_2} \right)^2$