A proton and an $$\alpha$$-particle (with their masses in the ratio of 1 : 4 and charges in the ratio of 1 : 2) are accelerated from rest through a potential difference V. If a uniform magnetic field (B) is set up perpendicular to their velocities, the ratio of the radii $$r_p : r_{\alpha}$$ of the circular paths described by them will be:

We begin by recalling the motion of a charged particle in a uniform magnetic field. When a particle of mass $$m$$ and charge $$q$$ enters a field $$B$$ with velocity $$v$$ perpendicular to the field, it moves in a circle of radius

$$r=\dfrac{mv}{qB}\,.$$

Next, because each particle starts from rest and is accelerated through the same potential difference $$V$$, the work-energy theorem gives the kinetic energy gained:

$$qV=\dfrac12\,mv^{2}\;.$$

Solving this energy equation for the speed, we obtain

$$v=\sqrt{\dfrac{2qV}{m}}\;.$$

Substituting this value of $$v$$ into the formula for the radius, we arrive at

$$r=\dfrac{m}{qB}\,\sqrt{\dfrac{2qV}{m}} =\dfrac{1}{B}\,\sqrt{\dfrac{2mV}{q}}\;.$$

Thus, for any particle accelerated through the same potential difference $$V$$ and entering the same magnetic field $$B$$, the radius is proportional to $$\sqrt{\dfrac{m}{q}}$$.

Let us now evaluate this expression for the two given particles.

For the proton: its mass is $$m_{p}=m$$ and its charge is $$q_{p}=e$$. Therefore

$$r_{p}=\dfrac1B\,\sqrt{\dfrac{2mV}{e}}\;.$$

For the $$\alpha$$-particle: its mass is $$m_{\alpha}=4m$$ and its charge is $$q_{\alpha}=2e$$. Hence

$$r_{\alpha}=\dfrac1B\,\sqrt{\dfrac{2(4m)V}{2e}} =\dfrac1B\,\sqrt{\dfrac{8mV}{2e}} =\dfrac1B\,\sqrt{\dfrac{4mV}{e}} =\dfrac1B\;\;2\,\sqrt{\dfrac{mV}{e}}\;.$$

We now write the ratio of the two radii:

$$\dfrac{r_{p}}{r_{\alpha}} =\dfrac{\displaystyle \dfrac1B\,\sqrt{\dfrac{2mV}{e}}} {\displaystyle \dfrac1B\,2\,\sqrt{\dfrac{mV}{e}}} =\dfrac{\sqrt{2}}{2} =\dfrac1{\sqrt{2}}\;.$$

This gives

$$r_{p}:r_{\alpha}=1:\sqrt{2}\;.$$

Hence, the correct answer is Option B.