Question

A proton and an alpha particle both enter a region of uniform magnetic field $B$, moving at right angles to the field $B$. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is $1\,MeV$, the energy acquired by the alpha particle will be

• A. $1.5\, MeV$
• B. $1\, MeV$
• C. $4 \,MeV$
• D. $0.5\, MeV$

Answer 1

The Correct Option is B

To find the energy acquired by the alpha particle, we need to analyze the motion of both the proton and the alpha particle under the influence of a uniform magnetic field. Both particles move in circular orbits due to the magnetic force acting as the centripetal force. The radius of the circular motion is given by the formula:

R = \frac{mv}{qB}

Where:

• m is the mass of the particle,
• v is the velocity of the particle,
• q is the charge of the particle, and
• B is the magnetic field.

Since the radii of both particles' orbits are equal, we can equate their radii:

\frac{m_p v_p}{q_p} = \frac{m_{\alpha} v_{\alpha}}{q_{\alpha}}

Where the subscripts p and \alpha refer to the proton and the alpha particle, respectively.

Given that the proton has a kinetic energy of 1 \, \text{MeV}, the velocity of the proton is:

K_p = \frac{1}{2} m_p v_p^2

Solving for v_p:

v_p = \sqrt{\frac{2K_p}{m_p}}

Substituting for v_p in terms of K_p in the radii equation:

\frac{m_p \sqrt{\frac{2K_p}{m_p}}}{q_p} = \frac{m_{\alpha} v_{\alpha}}{q_{\alpha}}

And rearranging gives:

v_{\alpha} = \frac{q_{\alpha}}{m_{\alpha}} \frac{v_p q_p}{m_p}

Considering that an alpha particle consists of 2 protons and 2 neutrons (with mass approximately 4 times that of a proton and charge twice that of a proton), the relationship becomes:

v_{\alpha} = \frac{2q_p}{4m_p} v_p = \frac{1}{2} v_p

The kinetic energy for the alpha particle K_{\alpha} is given by:

K_{\alpha} = \frac{1}{2} m_{\alpha} v_{\alpha}^2

Substituting m_{\alpha} = 4m_p and v_{\alpha} = \frac{1}{2} v_p:

K_{\alpha} = \frac{1}{2} \cdot 4m_p \cdot \left(\frac{1}{2} v_p\right)^2 = \frac{1}{2} \cdot 4m_p \cdot \frac{1}{4} v_p^2

K_{\alpha} = m_p v_p^2 = 2K_p

As the proton's kinetic energy was 1 MeV, the alpha particle's energy is also:

K_{\alpha} = 1 \, \text{MeV}

Hence, the energy acquired by the alpha particle is 1 \, \text{MeV}. Thus, the correct answer is:

• 1 \, \text{MeV}