Question

A proton and an alpha-particle moving with the same velocity enter a uniform magnetic field with their velocities perpendicular to the magnetic field. The ratio of radii of their circular paths is

• A. 1 : 4
• B. 4 : 1
• C. 1 : 2
• D. 2 : 1

Answer 1

The Correct Option is C

Step 1: Recall the formula for the radius of a charged particle moving in a magnetic field: 

When a charged particle with mass \(m\), charge \(q\), and velocity \(v\) enters a uniform magnetic field \(B\) perpendicular to the velocity, the radius (\(r\)) of its circular path is given by:

\[ r = \frac{mv}{qB} \]

Step 2: Define parameters clearly:

We have two particles:

• Proton: mass \(m_p\), charge \(+e\)
• Alpha-particle (He nucleus): mass \(4m_p\), charge \(+2e\)

Given: Both have the same velocity \(v\).

Step 3: Calculate radii separately:

- Radius for proton (\(r_p\)): \[ r_p = \frac{m_p v}{e B} \]

- Radius for alpha particle (\(r_\alpha\)): \[ r_\alpha = \frac{4 m_p v}{2 e B} = \frac{2 m_p v}{e B} \]

Step 4: Compute the ratio of radii:

\[ \frac{r_p}{r_\alpha} = \frac{\frac{m_p v}{e B}}{\frac{2 m_p v}{e B}} = \frac{1}{2} \]

Thus, the ratio of radii (proton : alpha-particle) is 1 : 2.