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 $$\left(\frac{r_S}{r_{\alpha}} \right)$$ is ___.

After acceleration through the same potential difference $$V$$, each ion gains kinetic energy equal to its charge times the potential:

For any ion with charge $$q$$ and mass $$m$$: $$\text{K.E.}=qV=\tfrac12 m v^{2} \; \Rightarrow \; v=\sqrt{\frac{2qV}{m}}$$

When the velocity is perpendicular to a uniform magnetic field $$B$$, the magnetic (Lorentz) force provides the necessary centripetal force:

$$\frac{mv^{2}}{r}=qvB \;\;\Rightarrow\;\; r=\frac{mv}{qB}$$

Substituting the expression for $$v$$:

$$r=\frac{m}{qB}\,\sqrt{\frac{2qV}{m}} \;=\;\frac1B\,\sqrt{\frac{2mV}{q}}$$

Thus the orbit radius depends only on the factor $$\sqrt{\dfrac{m}{q}}$$ (since $$2V/B^{2}$$ is common to both ions).
Therefore,

$$\frac{r_{S}}{r_{\alpha}} =\sqrt{\frac{m_{S}/q_{S}}{m_{\alpha}/q_{\alpha}}} =\sqrt{\frac{m_{S}\,q_{\alpha}}{m_{\alpha}\,q_{S}}}$$

Given data:
$$m_{\alpha}=4\;\text{amu}, \quad q_{\alpha}=+2e$$
$$m_{S}=32\;\text{amu}, \quad q_{S}=+e$$

Substituting:

$$\frac{r_{S}}{r_{\alpha}} =\sqrt{\frac{32\times 2e}{4\times e}} =\sqrt{\frac{64}{4}} =\sqrt{16}=4$$

Hence, the sulfur ion’s orbit is four times larger than that of the $$\alpha$$-particle.

Final answer: 4