A proton and an alpha particle moving with equal kinetic energies enter perpendicularly into a magnetic field. If $r_p$ and $r_α$ are the respective radii of their circular paths, what will be the ratio $r_p/r_α$?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1 : 1
The radius of a charged particle moving perpendicular to a magnetic field is:
$r = \frac{mv}{qB}$
Kinetic energy $K = \frac{1}{2} mv^2$ → $v = \sqrt{\frac{2K}{m}}$
Substitute $v$ into $r$:
$r = \frac{m \sqrt{2K/m}}{qB} = \frac{\sqrt{2 m K}}{q B}$
For proton:
$r_p = \frac{\sqrt{2 m_p K}}{q_p B}$
For alpha particle ($q_\alpha = 2 e$, $m_\alpha = 4 m_p$):
$r_\alpha = \frac{\sqrt{2 m_\alpha K}}{q_\alpha B} = \frac{\sqrt{2 (4 m_p) K}}{2 e B} = \frac{\sqrt{8 m_p K}}{2 e B} = \frac{\sqrt{2 m_p K} \cdot \sqrt{4}}{2 e B} = \frac{2 \sqrt{2 m_p K}}{2 e B} = \frac{\sqrt{2 m_p K}}{e B}$
Therefore, ratio:
$\frac{r_p}{r_\alpha} = \frac{\sqrt{2 m_p K}/(e B)}{\sqrt{2 m_p K}/(e B)} = 1$