In a nuclear experiment a proton with kinetic energy 1.0 MeV moves in a circular path in a unifo

step 1 So in a nuclear experiment there is a proton which has a kinetic energy Let us write that as E h

In a nuclear experiment a proton with kinetic energy 1.0 MeV...

In a nuclear experiment a proton with kinetic energy $1.0 \mathrm{MeV}$ moves in a circular path in a uniform magnetic field. What energy must (a) an alpha particle $(q=+2 e, m=4.0 \mathrm{u})$ and (b) a deuteron $(q=+e, m=2.0 \mathrm{u})$ have if they are to circulate in the same circular path?

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Key Concepts

Lorentz Force and Circular Motion

In a uniform magnetic field, any charged particle experiences the Lorentz force, which is perpendicular to its velocity. This force acts as the centripetal force that causes the particle to move in a circular path. The radius of this path is determined by the equation r = mv/(qB), where m is the mass, v is the speed, q is the charge, and B is the magnetic field strength.

Relationship between Kinetic Energy and Momentum

For a non?relativistic particle, the kinetic energy is given by K = ½ mv², and the momentum is p = mv. This leads to the relationship p = ?(2mK). When analyzing circular motion in a magnetic field, this relation allows one to connect the kinetic energy with the radius of the trajectory, which depends on the momentum of the particle.

Charge-to-Mass Ratio

The path radius in a magnetic field is dependent on the charge-to-mass ratio (q/m) of the particle. For two particles to have the same circular motion under the same magnetic field, their kinetic energies must adjust to account for differences in their q/m ratios. Essentially, to maintain the same radius, a particle with a different q or m than a reference particle needs a specific kinetic energy such that ?(2mK)/q remains constant.

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