In a mass spectrometer, a singly-charged particle (charge e) has a speed of 1.0 × 10⁶ m/s and enters a uniform magnetic field of 0.20 T. The radius of the circular orbit of the particle is 0.020 m. What is the mass of this particle? (e = 1.60 × 10^-19C) A) 3.2 × 10^-28 kg B) 6.4 × 10^-28 kg C) 1.7 × 10^-27 kg D) 4.5 × 10^-27 kg

Question

Quizplus · Accepted Answer

The radius of the circular path in a magnetic field is given by r = mv/(qB). Solving for mass (m), we have m = rqB/v. Substituting the given values: r = 0.020 m, q = 1.60 × 10^-19 C, B = 0.20 T, and v = 1.0 × 10^6 m/s, we find m = (0.020 m)(1.60 × 10^-19 C)(0.20 T) / (1.0 × 10^6 m/s) = 6.4 × 10^-28 kg.

In a Mass Spectrometer, a Singly-Charged Particle (Charge E) Has