Question

An $\alpha$-particle (mass $4 \,amu$ ) and a singly charged sulfur ion (mass $32\, amu$ ) are initially at rest. They are accelerated through a potential $V$ and then allowed to pass into a region of uniform magnetic field which is normal to the velocities of the particles. Within this region, the $\alpha$-particle and the sulfur ion move in circular orbits of radii $r_{\alpha}$ and $r_{s}$, respectively. The ratio \(\left(\frac{r_{s} }{ r_{a}}\right)\) is _____

Answer 1

Correct Answer: 4

Step-by-step Solution:

Concept: When charged particles are accelerated through a potential difference \(V\), they gain kinetic energy equal to the work done by the electric field:

\[ \text{K.E.} = qV \Rightarrow \frac{1}{2}mv^2 = qV \Rightarrow v = \sqrt{\frac{2qV}{m}} \]

In a magnetic field \(B\) perpendicular to the velocity, the radius of circular motion is given by: \[ r = \frac{mv}{qB} \Rightarrow r = \frac{m}{qB} \cdot \sqrt{\frac{2qV}{m}} = \sqrt{\frac{2mV}{q}} \cdot \frac{1}{B} \] Hence, \[ r \propto \sqrt{\frac{m}{q}} \]

Step 1: Write mass and charge for each particle

• α-particle:
  • Mass \( m_\alpha = 4 \, amu \)
  • Charge \( q_\alpha = 2e \)
• Singly charged sulfur ion:
  • Mass \( m_s = 32 \, amu \)
  • Charge \( q_s = 1e \)

Step 2: Take ratio of radii

\[ \frac{r_s}{r_\alpha} = \frac{ \sqrt{ \frac{m_s}{q_s} } }{ \sqrt{ \frac{m_\alpha}{q_\alpha} } } = \sqrt{ \frac{m_s q_\alpha}{m_\alpha q_s} } \] \[ = \sqrt{ \frac{32 \cdot 2}{4 \cdot 1} } = \sqrt{ \frac{64}{4} } = \sqrt{16} = 4 \]

Final Answer: 4