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An alpha particle can be produced in certain radioactive decays of nuclei and consists of two protons and two neutrons. The particle has a charge of q = +2e and a mass of 4.00 u, where u is the atomic mass unit, with 1 u = 1.661x10^-27 kg. Suppose an alpha particle travels in a circular path of radius 4.50 cm in a uniform magnetic field with B = 1.20 T. Calcuate (a) its speed, (b) its period of revolution, (c) its kinetic energy, and (d) the potential difference through which it would have to be accelerated to achieve this energy.

The book gives answers to this problem:
(a) = 2.6x10^6 m/s
(b) = 0.109 micro s
(c) = 140 MeV
(d) = 70.0 kV

User The Condor
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Final answer:

The speed of the alpha particle is 2.6x10^6 m/s. The period of revolution is 0.109 microseconds. The kinetic energy is 140 MeV and the potential difference required is 70.0 kV.

Step-by-step explanation:

To calculate the speed of the alpha particle, we can use the formula for the centripetal force exerted on a charged particle in a magnetic field. In this case, the centripetal force is provided by the magnetic force. We have the formula F = qvB, where F is the force, q is the charge, v is the velocity, and B is the magnetic field strength.

To find the speed of the alpha particle, we need to rearrange the formula to solve for v. We have qvB = mv^2/r, where m is the mass of the alpha particle and r is the radius of the circular path.

(a) v = (qBr/m)^0.5

Substituting the given values q = +2e, B = 1.20 T, r = 4.50 cm = 0.045 m, and m = 4.00 u = 4.00 x 1.661x10^-27 kg:

v = [(2e)(1.20 T)(0.045 m)]^0.5 / (4.00 x 1.661x10^-27 kg) = 2.6x10^6 m/s

Therefore, the speed of the alpha particle is 2.6x10^6 m/s.

(b) The period of revolution of the alpha particle can be calculated using the formula T = 2πr/v, where T is the period, r is the radius, and v is the velocity.

T = 2π(0.045 m) / (2.6x10^6 m/s) = 0.109 micro s

Therefore, the period of revolution of the alpha particle is 0.109 micro s.

(c) The kinetic energy of the alpha particle can be calculated using the formula KE = 0.5mv^2, where KE is the kinetic energy, m is the mass, and v is the velocity.

KE = 0.5(4.00 x 1.661x10^-27 kg) (2.6x10^6 m/s)^2 = 140 MeV

Therefore, the kinetic energy of the alpha particle is 140 MeV.

(d) The potential difference needed to accelerate the alpha particle can be calculated using the formula KE = qV, where KE is the kinetic energy, q is the charge, and V is the potential difference.

V = KE/q = (140 MeV) / (2e) = 70.0 kV

Therefore, the potential difference through which the alpha particle would have to be accelerated to achieve this energy is 70.0 kV.

User Paul Slocum
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