Question

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

• A. The momentum of the second ion is twice the momentum of the first ion.
• B. The velocity of the first ion is twice the velocity of the second ion.
• C. The radius of the first ion's path is 3/4 times the radius of the second ion's path.

Answer 1

Answer: (A), (B), (C)

All of the above

Answer 2

Let $m_1 = 4 \text{ amu}$ and $m_2 = 16 \text{ amu}$ be the masses of the two ions, and let $q_1 = +2e$ and $q_2 = +3e$ be their respective charges. Let $K$ be the kinetic energy of both ions.

1. Momentum comparison: Since $K = \frac{p^2}{2m}$, we have $p = \sqrt{2mK}$. Therefore,

$\frac{p_1}{p_2} = \frac{\sqrt{2m_1K}}{\sqrt{2m_2K}} = \sqrt{\frac{m_1}{m_2}} = \sqrt{\frac{4}{16}} = \frac{1}{2}$

Thus, $p_2 = 2p_1$. The momentum of the second ion is twice the momentum of the first ion.

2. Velocity comparison: Since $K = \frac{1}{2}mv^2$, we have $v = \sqrt{\frac{2K}{m}}$. Therefore,

$\frac{v_1}{v_2} = \frac{\sqrt{\frac{2K}{m_1}}}{\sqrt{\frac{2K}{m_2}}} = \sqrt{\frac{m_2}{m_1}} = \sqrt{\frac{16}{4}} = 2$

Thus, $v_1 = 2v_2$. The velocity of the first ion is twice the velocity of the second ion.

3. Radius comparison: The radius of the circular path in a magnetic field is given by $r = \frac{mv}{qB} = \frac{p}{qB}$. Therefore,

$\frac{r_1}{r_2} = \frac{p_1/q_1B}{p_2/q_2B} = \frac{p_1}{p_2} \cdot \frac{q_2}{q_1} = \frac{1}{2} \cdot \frac{3e}{2e} = \frac{3}{4}$

Thus, $r_1 = \frac{3}{4}r_2$. The radius of the first ion's path is 3/4 times the radius of the second ion's path.

All three relationships derived above are correct conclusions.