Alpha particles, each having a charge of +2e and a mass of 6.64 ×10^-27 kg, are accelerated in a uniform $0.50 \mathrm {~T}$ magnetic field to a final orbit radius of $0.50 \mathrm {~m}$ The field is perpendicular to the velocity of the particles. What is the kinetic energy of an alpha particle in the final orbit? (1 eV = 1.60 × 10^-19 J, e = 1.60 × 10^-19C) A) 3.0 MeV B) 2.6 MeV C) 3.4 MeV D) 3.9 MeV E) 4.3MeV

Question

Quizplus · Accepted Answer

The correct answer is A because the kinetic energy of an alpha particle in the final orbit is 3.0 MeV. This can be calculated using the formula: KE = (1/2)mv^2 = (1/2)(6.64 ×10^-27 kg)(v^2), where v is the velocity of the alpha particle. The velocity can be calculated using the formula: v = (qB/m)r, where q is the charge of the alpha particle, B is the magnetic field strength, and r is the radius of the orbit. Substituting the given values into these formulas, we get: v = (2e)(0.50 T)/(6.64 ×10^-27 kg)(0.50 m) = 3.0 ×10^7 m/s and KE = (1/2)(6.64 ×10^-27 kg)(3.0 ×10^7 m/s)^2 = 3.0 MeV.

Alpha Particles, Each Having a Charge of +2e and a Mass