Proton and Alpha Particle in a Magnetic Field - Detailed Solution

Proton and Alpha Particle in a Magnetic Field

Question: A proton carrying 1 MeV kinetic energy is moving in a circular path of radius \( R \) in a uniform magnetic field. What should be the kinetic energy of an alpha particle to describe a circle of the same radius in the same field?

(a) 2 MeV

(b) 0.5 MeV

(c) 1 MeV

(d) 4 MeV

Detailed Step-by-Step Explanation

Step 1: Magnetic Radius Formula
A charged particle moving with momentum \( p \) in a uniform magnetic field \( B \) perpendicular to its velocity follows a circular path with radius given by: \[ r = \frac{p}{qB}, \] where \( q \) is the charge of the particle.

Step 2: Relate Momentum and Kinetic Energy
The kinetic energy \( K \) of a particle is related to its momentum by: \[ K = \frac{p^2}{2m}. \]

Step 3: Equal Radius Condition
For both the proton and the alpha particle to have the same radius in the same magnetic field, we require: \[ \frac{p_p}{q_p} = \frac{p_\alpha}{q_\alpha}. \] For a proton, \( q_p = e \). For an alpha particle, \( q_\alpha = 2e \). Thus: \[ \frac{p_p}{e} = \frac{p_\alpha}{2e} \quad \Rightarrow \quad p_\alpha = 2p_p. \]

Step 4: Relate Kinetic Energies
Let \( K_p = 1 \) MeV be the kinetic energy of the proton and \( K_\alpha \) that of the alpha particle. We have: \[ p_p^2 = 2m_p K_p, \] and \[ p_\alpha^2 = 2m_\alpha K_\alpha. \] Since \( p_\alpha = 2p_p \), then: \[ (2p_p)^2 = 4p_p^2 = 2m_\alpha K_\alpha. \]

Step 5: Substitute Mass of Alpha Particle
The mass of an alpha particle is approximately \( m_\alpha \approx 4m_p \). Substituting: \[ 4p_p^2 = 2(4m_p) K_\alpha = 8m_p K_\alpha. \] Thus: \[ K_\alpha = \frac{4p_p^2}{8m_p} = \frac{p_p^2}{2m_p}. \]

Notice that: \[ \frac{p_p^2}{2m_p} = K_p = 1\,\text{MeV}. \] Therefore, the kinetic energy of the alpha particle is: \[ K_\alpha = 1\,\text{MeV}. \]

Final Answer: The kinetic energy acquired by the alpha particle is \( 1\,\text{MeV} \), which corresponds to option (c).