Two ions of masses 4 amu and 16 amu have charges +2e and +3e respectively. These ions pass through the region of the constant perpendicular magnetic field. The kinetic energy of both ions is the same. Then:

We have two ions that both enter a region where the magnetic field $$\vec B$$ is uniform and perpendicular to their velocities. Whenever a charged particle moves perpendicular to a magnetic field, it describes a circular path. The required formula for the radius of that circular path is first recalled:

For a particle of mass $$m$$, charge $$q$$ and speed $$v$$ in a magnetic field of magnitude $$B$$ (with velocity perpendicular to the field), the magnetic force provides the necessary centripetal force. Hence

$$q\,v\,B = \dfrac{m v^{2}}{r}.$$

Solving this equation for the radius $$r$$ gives

$$r = \dfrac{m\,v}{q\,B} \;.$$

In the present question, both ions possess the same kinetic energy. Let us denote that common kinetic energy by $$K$$. The kinetic energy of any particle is related to its speed by the well-known relation

$$K = \dfrac{1}{2} m v^{2}.$$

From this we can express the speed $$v$$ of each ion in terms of its mass and the common kinetic energy:

$$v = \sqrt{\dfrac{2K}{m}}.$$

Now we substitute this expression for $$v$$ into the radius formula. Performing the substitution, we get

$$r = \dfrac{m}{q\,B}\;\sqrt{\dfrac{2K}{m}}.$$

By taking the square root of $$m$$ in the numerator, the algebraic simplification proceeds as follows:

$$r = \dfrac{\sqrt{m}\;\sqrt{m}}{q\,B}\;\sqrt{\dfrac{2K}{m}} = \dfrac{\sqrt{m}\;}{q\,B}\;\sqrt{2K} = \dfrac{\sqrt{2K}}{B}\;\dfrac{\sqrt{m}}{q}.$$

The magnetic field $$B$$ and the factor $$\sqrt{2K}$$ are identical for both ions, so the only quantity that decides the relative radii is the ratio $$\dfrac{\sqrt{m}}{q}$$.

Let us calculate this ratio for each ion separately.

Ion 1 (lighter ion)
Mass $$m_{1}=4\ \text{amu},\qquad q_{1}=+2e.$$

$$\dfrac{\sqrt{m_{1}}}{q_{1}} = \dfrac{\sqrt{4}}{2e} = \dfrac{2}{2e} = \dfrac{1}{e}.$$

Ion 2 (heavier ion)
Mass $$m_{2}=16\ \text{amu},\qquad q_{2}=+3e.$$

$$\dfrac{\sqrt{m_{2}}}{q_{2}} = \dfrac{\sqrt{16}}{3e} = \dfrac{4}{3e}.$$

We now compare these two numerical factors:

$$\dfrac{\sqrt{m_{2}}}{q_{2}} = \dfrac{4}{3e}\;>\;\dfrac{1}{e} = \dfrac{\sqrt{m_{1}}}{q_{1}}.$$

The radius of curvature is directly proportional to this factor, so

$$r_{2} > r_{1}.$$

Because the heavier ion (Ion 2) travels in a circle of larger radius, it is bent less sharply by the magnetic field. Conversely, the lighter ion (Ion 1) follows a circle of smaller radius, meaning its trajectory bends more.

In ordinary language, a smaller radius corresponds to greater deflection; a larger radius corresponds to less deflection. Therefore the lighter ion is deflected more, and the heavier ion is deflected less.

Hence, the correct answer is Option B.