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Step-by-Step Solution
Step 1: Identify the Known Information
• A deuteron (charge = e, mass = 2mp) and an alpha particle (charge = 2e, mass = 4mp) have the same kinetic energy K.
• Both particles enter a uniform magnetic field perpendicularly.
Step 2: Recall the Formula for the Radius of the Circular Path
When a charged particle of charge $q$, mass $m$, and velocity $v$ moves perpendicularly to a magnetic field $B$, the radius $r$ of the circular path is
$r = \frac{mv}{Bq}$.
Step 3: Express Particle Velocity in Terms of the Kinetic Energy
The kinetic energy of each particle is
$K = \frac{1}{2} mv^2 \implies mv^2 = 2K \implies v = \sqrt{\frac{2K}{m}}.$
Step 4: Substitute Velocity into the Radius Formula
Substitute $v = \sqrt{\frac{2K}{m}}$ into $r = \frac{mv}{Bq}$:
$r = \frac{m \sqrt{\frac{2K}{m}}}{Bq} \;=\; \frac{\sqrt{2Km}}{Bq}.$
Step 5: Conclude the Proportionality and Form the Ratio
From $r = \frac{\sqrt{2Km}}{Bq}$, we see that
$r \propto \frac{\sqrt{m}}{q},
since $K$ and $B$ are the same for both particles. Thus, for the deuteron (subscript d) and the alpha particle (subscript α):
$\frac{r_d}{r_\alpha} = \frac{\sqrt{m_d}}{\sqrt{m_\alpha}} \cdot \frac{q_\alpha}{q_d}.$
Step 6: Substitute Masses and Charges
For a deuteron: $m_d = 2m_p$, $q_d = e$.
For an alpha particle: $m_\alpha = 4m_p$, $q_\alpha = 2e.$
Hence,
$\frac{r_d}{r_\alpha}
= \sqrt{\frac{2m_p}{4m_p}} \times \frac{2e}{e}
= \sqrt{\frac{2}{4}} \times 2
= \frac{\sqrt{2}}{2} \times 2
= \sqrt{2}.$
Step 7: Final Answer
Therefore, the ratio of the radii of the circular paths is
$\boxed{\sqrt{2}.}$
