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 E) 3.1 × 10-^-31 kg

Question

Quizplus · Accepted Answer

The mass of the particle can be found using the formula for the radius $r$ of the circular path of a charged particle in a magnetic field: $r = \frac{mv}{qB}$, where $m$ is the mass of the particle, $v$ is its velocity, $q$ is its charge, and $B$ is the magnetic field strength. Rearranging for $m$, we get $m = \frac{rqB}{v}$. Substituting the given values ($r = 0.020 \, \text{m}$, $q = 1.60 \times 10^{-19} \, \text{C}$, $B = 0.20 \, \text{T}$, $v = 1.0 \times 10^{6} \, \text{m/s}$), we find $m = \frac{0.020 \times 1.60 \times 10^{-19} \times 0.20}{1.0 \times 10^{6}} = 6.4 \times 10^{-28} \, \text{kg}$.

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