Charged Particles in a Magnetic Field
Question: A proton and an alpha particle both enter a region of uniform magnetic field \( B \) at right angles to the field. If the radii of their circular orbits are equal and the kinetic energy acquired by the proton is 1 MeV, what is the kinetic energy acquired by the alpha particle? (2015)
Detailed Step-by-Step Explanation
Step 1: Expression for the Radius in a Magnetic Field
When a charged particle of charge \( q \) moves with momentum \( p \) perpendicular to a uniform magnetic field \( B \), it follows a circular path with radius:
\[
r = \frac{p}{qB}.
\]
Step 2: Equal Radii Condition
Since the proton and the alpha particle have the same radius, we equate:
\[
\frac{p_p}{q_pB} = \frac{p_\alpha}{q_\alpha B}.
\]
This implies:
\[
p_\alpha = \frac{q_\alpha}{q_p}\,p_p.
\]
For a proton, \( q_p = e \); for an alpha particle, \( q_\alpha = 2e \). Therefore:
\[
p_\alpha = 2p_p.
\]
Step 3: Kinetic Energy Relations
The kinetic energy of a particle is given by:
\[
K = \frac{p^2}{2m}.
\]
For the proton, the kinetic energy is:
\[
K_p = \frac{p_p^2}{2m_p} = 1\,\text{MeV}.
\]
For the alpha particle:
\[
K_\alpha = \frac{p_\alpha^2}{2m_\alpha}.
\]
Substituting \( p_\alpha = 2p_p \) gives:
\[
K_\alpha = \frac{(2p_p)^2}{2m_\alpha} = \frac{4p_p^2}{2m_\alpha} = \frac{2p_p^2}{m_\alpha}.
\]
Step 4: Relate the Masses
The mass of an alpha particle is approximately four times the mass of a proton:
\[
m_\alpha \approx 4m_p.
\]
Thus, the kinetic energy for the alpha particle becomes:
\[
K_\alpha = \frac{2p_p^2}{4m_p} = \frac{p_p^2}{2m_p}.
\]
Notice that: \[ \frac{p_p^2}{2m_p} = K_p = 1\,\text{MeV}. \] Therefore: \[ K_\alpha = 1\,\text{MeV}. \]
Final Answer: The kinetic energy acquired by the alpha particle is \( 1\,\text{MeV} \), which corresponds to option (b).
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