In a certain cyclotron,a proton of mass 1.67 × 10^-27 kg moves in a circle of diameter 1.6 m with an angular speed of 2.0 × 10⁶ rad/s.What is the angular momentum of the proton? A)1.3 × 10^-21 kg ∙ m²/s B)1.8 × 10^-21 kg ∙ m²/s C)2.1 × 10^-21 kg ∙ m²/s D)3.2 × 10^-21 kg ∙ m²/s

Question

Quizplus · Accepted Answer

The angular momentum of a particle rotating in a circular orbit is given by:

L = mvr

where m is the mass of the particle, v is its speed, and r is the radius of the orbit.

In this case, the diameter of the orbit is given as 1.6 m, so the radius is r = 0.8 m. The speed of the proton can be found from its angular speed:

v = ωr = (2.0 × 10^16 rad/s)(0.8 m) = 1.6 × 10^16 m/s

Substituting the mass of the proton m = 1.67 × 10^-27 kg, we get:

L = (1.67 × 10^-27 kg)(1.6 × 10^16 m/s)(0.8 m) = 2.13 × 10^-21 kg·m^2/s

Therefore, the angular momentum of the proton is 2.1 × 10^-21 kg·m^2/s, which corresponds to choice C.

In a Certain Cyclotron,a Proton of Mass 1