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An electron, a proton, a deuteron and an...

An electron, a proton, a deuteron and an alpha particle, each having the same speed are in a region of constant magnetic field perpendicular to the direction of the velocities of the particles. The radius of the circular orbits of these particles are respectively `R_(e), R_(p), R_(d)` and `R_(alpha)` It follows that

A

`R_e = R_p`

B

`R_p = R_d`

C

`R_d = R_(alpha)`

D

`R_p = R_alpha`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to determine the relationship between the radii of the circular orbits of an electron, a proton, a deuteron, and an alpha particle when they are moving in a magnetic field with the same speed. ### Step-by-Step Solution: 1. **Understanding the Motion in a Magnetic Field**: When a charged particle moves in a magnetic field, it experiences a magnetic force that acts as a centripetal force, causing it to move in a circular path. The magnetic force \( F \) on a charged particle is given by: \[ F = qvB \] where \( q \) is the charge of the particle, \( v \) is its velocity, and \( B \) is the magnetic field strength. 2. **Equating Forces**: The magnetic force provides the centripetal force necessary for circular motion, which is given by: \[ F = \frac{mv^2}{r} \] where \( m \) is the mass of the particle and \( r \) is the radius of the circular path. Setting these two expressions for force equal gives: \[ qvB = \frac{mv^2}{r} \] 3. **Solving for the Radius**: Rearranging the equation to solve for the radius \( r \): \[ r = \frac{mv}{qB} \] 4. **Analyzing Charge-to-Mass Ratios**: Since all particles have the same speed \( v \) and are in the same magnetic field \( B \), the radius \( r \) depends on the charge-to-mass ratio \( \frac{q}{m} \): \[ r \propto \frac{m}{q} \] 5. **Calculating Charge and Mass for Each Particle**: - **Electron**: - Charge \( q_e = -e \) (magnitude \( e \)) - Mass \( m_e \approx 9.11 \times 10^{-31} \) kg - Charge-to-mass ratio \( \frac{q_e}{m_e} = \frac{e}{m_e} \) - **Proton**: - Charge \( q_p = +e \) - Mass \( m_p \approx 1.67 \times 10^{-27} \) kg - Charge-to-mass ratio \( \frac{q_p}{m_p} = \frac{e}{m_p} \) - **Deuteron**: - Charge \( q_d = +e \) - Mass \( m_d \approx 2 \times 1.67 \times 10^{-27} \) kg (roughly twice the proton mass) - Charge-to-mass ratio \( \frac{q_d}{m_d} = \frac{e}{2m_p} \) - **Alpha Particle**: - Charge \( q_{\alpha} = +2e \) - Mass \( m_{\alpha} \approx 4 \times 1.67 \times 10^{-27} \) kg (roughly four times the proton mass) - Charge-to-mass ratio \( \frac{q_{\alpha}}{m_{\alpha}} = \frac{2e}{4m_p} = \frac{e}{2m_p} \) 6. **Comparing the Ratios**: - For the electron, \( \frac{q_e}{m_e} \) is much larger than for the proton, deuteron, and alpha particle. - For the proton and deuteron, the charge-to-mass ratios are the same. - For the alpha particle, it has the same charge-to-mass ratio as the deuteron. 7. **Conclusion**: The radii of the circular orbits are related as follows: \[ R_e > R_p = R_d = R_{\alpha} \] Therefore, the correct option is \( R_p = R_d = R_{\alpha} \). ### Final Answer: The relationship is \( R_e > R_p = R_d = R_{\alpha} \).
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