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Step-by-Step Solution
Step 1: Identify relevant physics concepts
When a charged particle is accelerated from rest through a potential difference $V$, it gains kinetic energy equal to $qV$, where $q$ is the charge of the particle.
Subsequently, when entering a uniform magnetic field $\mathbf{B}$ perpendicular to its velocity, the particle undergoes uniform circular motion.
The radius $r$ of this circular path depends on the mass $m$ of the particle, its charge $q$, and the magnetic field $B$.
Step 2: Write down the expression for velocity
The kinetic energy gained by a particle of mass $m$ and charge $q$ after being accelerated through potential difference $V$ is:
$$ \frac{1}{2} m v^2 = q V. $$
Solving for the speed $v$:
$$ v = \sqrt{\frac{2 q V}{m}}. $$
Step 3: Write down the expression for radius of circular motion
In a uniform magnetic field $B$, the radius $r$ of the circular path described by the particle is:
$$ r = \frac{m v}{q B}. $$
Substituting the expression for $v$, we get:
$$ r = \frac{m}{q B} \sqrt{\frac{2 q V}{m}} = \frac{\sqrt{2 m q V}}{q B}. $$
Step 4: Compare the radii for proton and $\alpha$-particle
Let us denote:
• $m_p$ and $q_p$ as the mass and charge of the proton, respectively.
• $m_\alpha$ and $q_\alpha$ as the mass and charge of the $\alpha$-particle, respectively.
From the problem statement:
$m_p : m_\alpha = 1 : 4$ and $q_p : q_\alpha = 1 : 2.$
Step 5: Calculate each radius
Using the formula
$$ r = \frac{\sqrt{2 m q V}}{q B}, $$
we find:
• Radius for the proton, $r_p$:
$$ r_p = \frac{\sqrt{2 \, m_p \, q_p \, V}}{q_p \, B}. $$
• Radius for the $\alpha$-particle, $r_\alpha$:
$$ r_\alpha = \frac{\sqrt{2 \, m_\alpha \, q_\alpha \, V}}{q_\alpha \, B}. $$
Step 6: Substitute numerical ratios and simplify
Let $m_p = 1$, $m_\alpha = 4$, $q_p = 1$, and $q_\alpha = 2$ (using the given ratios):
• $r_p = \dfrac{\sqrt{2 \cdot 1 \cdot 1 \cdot V}}{1 \cdot B} = \frac{\sqrt{2V}}{B}.$
• $r_\alpha = \dfrac{\sqrt{2 \cdot 4 \cdot 2 \cdot V}}{2 \cdot B}
= \dfrac{\sqrt{16V}}{2B}
= \dfrac{4\sqrt{V}}{2B}
= \dfrac{2\sqrt{V}}{B}.$
Step 7: Find the ratio of the radii
The ratio
$$ \frac{r_p}{r_\alpha}
= \frac{\sqrt{2V}/B}{2\sqrt{V}/B}
= \frac{\sqrt{2V}}{2\sqrt{V}}
= \frac{\sqrt{2}}{2}
= \frac{1}{\sqrt{2}}. $$
Hence,
$$ r_p : r_\alpha = 1 : \sqrt{2}. $$
Final Answer
The ratio of the radii of the circular paths (proton : $\alpha$-particle) is
$$ 1 : \sqrt{2}. $$
