An \alpha -particle (mass 4 amu) and a singly charged sulphur 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 rs, respectively. The ratio (rs/r \alpha ) is \underline\hspace2cm .

Magnetism PYQ — JEE Advanced 2021 · Shift 1 · Q55 with Solution

View written solutionFree

Correct answer: 4

For a charged particle entering a uniform magnetic field perpendicular to its velocity, $r = \frac{mv}{qB}$

If a particle starts from rest and is accelerated through potential $V$ , then $qV = \frac{1}{2}mv^2$ So, $v = \sqrt{\frac{2qV}{m}}$

$r = \frac{m}{qB}\sqrt{\frac{2qV}{m}}$

Simplifying, $r = \frac{1}{B}\sqrt{\frac{2mV}{q}}$

Thus, $r \propto \sqrt{\frac{m}{q}}$

For the $\alpha$ -particle:
- mass $m_\alpha = 4\,\text{amu}$
- charge $q_\alpha = 2e$
For the singly charged sulphur ion:
- mass $m_s = 32\,\text{amu}$
- charge $q_s = e$

Therefore, $\frac{r_s}{r_\alpha} = \sqrt{\frac{m_s/q_s}{m_\alpha/q_\alpha}}$

$\frac{r_s}{r_\alpha} = \sqrt{\frac{32/e}{4/(2e)}}$

$\frac{r_s}{r_\alpha} = \sqrt{\frac{32}{1}\cdot\frac{2}{4}}$

$\frac{r_s}{r_\alpha} = \sqrt{16} = 4$

$\boxed{4}$

The derived answer matches the stored correct answer.