1. Motion of a charged particle in a perpendicular magnetic field

When a charged particle enters a magnetic field perpendicular to its velocity, it moves in a circular path of radius

$r = \frac{mv}{qB}$

where:

• $m$ = mass of ion
• $v$ = speed of ion
• $q$ = charge of ion
• $B$ = magnetic field

The amount of deflection is greater for smaller radius and less for larger radius.

2. Use the condition of equal kinetic energy

Given that both ions have the same kinetic energy,

$K = \frac{1}{2}mv^2$

So,

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

Substitute into radius formula:

$r = \frac{m}{qB}\sqrt{\frac{2K}{m}} = \frac{\sqrt{2Km}}{qB}$

Thus,

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

3. Calculate for both ions

Ion 1 (lighter ion)

• Mass $m_1 = 4\,\text{amu}$
• Charge $q_1 = +2e$

$r_1 \propto \frac{\sqrt{4}}{2} = \frac{2}{2} = 1$

Ion 2 (heavier ion)

• Mass $m_2 = 16\,\text{amu}$
• Charge $q_2 = +3e$

$r_2 \propto \frac{\sqrt{16}}{3} = \frac{4}{3}$

So,

$r_2 > r_1$

The heavier ion has a larger radius, so it is deflected less. The lighter ion has a smaller radius, so it is deflected more.

4. Check options

• A: lighter ion will be deflected less than heavier ion ❌
• B: lighter ion will be deflected more than heavier ion ✅
• C: both ions will be deflected equally ❌
• D: no ion will be deflected ❌

5. Final answer

The correct option is

$\boxed{\text{B}}$