An 𝛼-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 𝛼-particle and the sulfur ion move in circular orbits of radii 𝑟𝛼 and 𝑟𝑆, respectively. The ratio (𝑟𝑆/𝑟𝛼) is ___.

 To determine the ratio 

$\frac{r_S}{r_{\alpha }}$ of the radii of the circular orbits of the sulfur ion ($S$) and the alpha particle ($\alpha$) in a uniform magnetic field, we need to follow these steps:

1. Understanding the Problem

• Particles:

  • Alpha particle ($\alpha$): mass $m_{\alpha }=4$ amu, charge $q_{\alpha }=+2e$

  • Singly charged sulfur ion ($S$): mass $m_S=32$ amu, charge $q_S=+e$

• Initial State: Both particles are at rest.

• Acceleration: Both particles are accelerated through a potential $V$.

• Magnetic Field: After acceleration, they enter a region with a uniform magnetic field $B$ perpendicular to their velocities.

• Motion: In the magnetic field, both particles move in circular orbits with radii $r_{\alpha }$ and $r_S$.

2. Calculating the Kinetic Energy Gained from the Potential

When a charged particle is accelerated through a potential $V$, it gains kinetic energy equal to the work done by the electric field:

$$K=qV$$

where:

• $K$ is the kinetic energy,

• $q$ is the charge of the particle,

• $V$ is the potential difference.

For the alpha particle:

$$K_{\alpha }=q_{\alpha }V=2eV$$

For the sulfur ion:

$$K_S=q_{S}V=eV$$

3. Relating Kinetic Energy to Velocity

The kinetic energy of a particle can also be expressed in terms of its mass and velocity:

$$K=\frac{1}{2}m{v}^2$$

Solving for velocity $v$:

$$v=\sqrt{\frac{2K}{m}}$$

For the alpha particle:

$$v_{\alpha }=\sqrt{\frac{2\times 2eV}{4\text{ amu}}}=\sqrt{\frac{4eV}{4\text{ amu}}}=\sqrt{\frac{eV}{\text{ amu}}}$$

For the sulfur ion:

$$v_S=\sqrt{\frac{2\times eV}{32\text{ amu}}}=\sqrt{\frac{2eV}{32\text{ amu}}}=\sqrt{\frac{eV}{16\text{ amu}}}$$

4. Determining the Radius of the Circular Orbit in a Magnetic Field

When a charged particle moves perpendicular to a uniform magnetic field, it experiences a centripetal force that causes it to move in a circular path. The radius $r$ of this path 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.

For the alpha particle:

$$r_{\alpha }=\frac{m_{\alpha }{v}_{\alpha }}{q_{\alpha }B}=\frac{4\text{ amu}\times \sqrt{\frac{eV}{\text{ amu}}}}{2eB}=\frac{4\sqrt{eV}}{2eB}=\frac{2\sqrt{eV}}{eB}$$

For the sulfur ion:

$$r_S=\frac{m_S{v}_S}{q_{S}B}=\frac{32\text{ amu}\times \sqrt{\frac{eV}{16\text{ amu}}}}{eB}=\frac{32\times \frac{\sqrt{eV}}{4}}{eB}=\frac{8\sqrt{eV}}{eB}$$

5. Calculating the Ratio $\frac{r_S}{r_{\alpha }}$

Now, we can find the ratio of the radii:

$$\frac{r_S}{r_{\alpha }}=\frac{\frac{8\sqrt{eV}}{eB}}{\frac{2\sqrt{eV}}{eB}}=\frac{8}{2}=4$$

6. Conclusion

The ratio of the radii of the circular orbits of the sulfur ion to the alpha particle is 4.

Final Answer: $\frac{r_S}{r_{\alpha }}=4$