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 1.55E+0 m radius perpendicular to a magnetic field of 2.70E-2 T. The kinetic energy of the particle is determined to be 6.70E-15 J. Identify the particle from its mass.

positron= 9.10E-31 kg

muon= 1.88E-28 kg

pion=2.50E-28 kg

kaon= 8.84E-28 kg

proton= 1.67E-27 kg

Dmeson= 3.35E-27 kg

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. Answer is case sensitive!!.

Explanation / Answer

Here, Radius ,r = mv/qB

Also, v = sqrt(2 * 6.70 * 10-15/m)

=>   1.55 = m * (sqrt(2 * 6.70 * 10-15/m)) / (1.6 * 10-19 * 0.027)

=>    0.06696 * 10-19 =   sqrt(m) * 11.5758 * 10-8

=>   m =   3.346 * 10-27 kg

=> particle is    Dmeson

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