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 2.18×10-1 m radius perpendicular to a magnetic field of 2.70×10-2 T. The kinetic energy of the particle is determined to be 2.35×10-15 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

Given data

Radius = R = 2.18* 10-1 m

magnetic field = B = 2.7* 10-2 T

Kinetic energy = K.E = 2.35 * 10-15 J

Mass = m = ?

SOlution

As particle is moving in circular arc and centripetal force will be provided by the magnetic force

Fc = Fm

mV2 / R = Bev

m = Be R / v ..............1

Here V = ?

By using Kinetic energy formula

K.E = 0.5 mV2

2.35 * 10-15 /0.5 V2 = m .......................2

comparing equation 1 and equ2

2.35* 10-15 / 0.5 V2 = BeR / v

V3 = 0.5BeR / 2.35* 10-15

V3 = 0.5(2.7*10-2)(1.67*10-19)(2.18*10-1) / 2.35*10-15

  =2.09 * 10-7

V = 1.27 *0.0046

=0.0058m/s

putting in equatio 1

m = BeR / V

m = 1.39* 10-10 Kg

According to our calcualtion there is no particle in above list matching to our answer.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.