In a high-energy physics experiment, a subnuclear particle moves in a circular a

Question

In a high-energy physics experiment, a subnuclear particle moves in a circular arc of 1.02 m radius perpendicular to a magnetic field of 2.70×10-2 T. The kinetic energy of the particle is determined to be 1.10×10-14 J. Identify the particle from its mass. The masses of the positron, pion, kaon, proton, muon and Dmeson are 9.10×10-31 kg, 2.50×10-28 kg, 8.84×10-28 kg, 1.67×10-27 kg, 1.88×10-28 kg and 3.35×10-27 kg, respectively. Assume that the particle is known to have a positive charge equal to the magnitude of the electron charge. Enter the name of the particle from the list given above.

Using this equation to solve this:
q v B = mv^2/r
mv = q B r
m^2 v^2 = [q B r]^2
2m [(1/2) mv^2] = [q B r]^2
m = mass = [q B r]^2 / (2 * kinetic energy)

Explanation / Answer

The magnitude of the vharge of the particle is q = 1.602*10-19 C The applied magnetic field is B = 2.7*10-2 T The radius of the circular path is R = 1.02 m The kinetic energy of the molecule is K = 1.1*10-14 J The mass of the charged particle is given by m = (BqR)2/2K m = [(2.7*10-2 T)(1.602*10-19 C)(1.02 m)]2/2(1.1*10-14 J) m = 8.847*10-28 kg By comparison of mass, the partilce should be KAON

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