In a high-energy physics experiment, a subnuclear particle moves in a circular a
ID: 1400463 • Letter: I
Question
In a high-energy physics experiment, a subnuclear particle moves in a circular arc of 3.09 m radius perpendicular to a magnetic field of 2.70×10-2 T. The kinetic energy of the particle is determined to be 5.34×10-14 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.
Explanation / Answer
As it is given that the charge on the particle is q = 1.6*10-19 C
therefore
qVB = mV2/ R
qBR = =mV where q is charge , B is magnetic field , R is radius of the circular path
on squaring both side
m2V2 = (qBR)2
we can write the above equation as follows
2m((1/2)mV2) = (qBR)2
2m K = (qBR)2 where K is kinetic energy
m = (qBR)2 / 2K
= (1.6*10-19*2.7*10-2*3.09)2 / (2*5.34*10-14) = 1.67*10-27 kg
therefore the particle will be the proton
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