An alpha particle can be produced in certain radioactive decays of nuclei and co

Question

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

I am stuck on part (a) where we determine the speed. I have been using the equation:

v = mr/qB ==> v = (6.644x10^-27 kg)(0.045 m) / (3.2x10^-19 C)(1.2 N/A*m)

where m = mass, r = radius, q = charge and B = magnetic field but I am not getting the answer the book gives. Am I using the wrong equation or values?

Explanation / Answer

You are using the correct equation, but you have it written incorrectly

The equation is r = mv/qB. therefore v = rqB/m

Part A)

v = (.045)(2)(1.6 X 10-19)(1.2)/(4)(1.661 X 10-27)

v = 2.6 X 106 m/s

Part B)

Period of revolution is distance over velocity for one circle

T = (2)(.045)/(2.6 X 106)

T = 1.09 X 10-7 s   which is .109 s

Part C)

KE = .5mv2

KE = (.5)(4)(1.661 X 10-27)(2.6 X 106)2

KE = 2.25 X 10-14 J  

Since there are 1.602 X 10-19 J/eV, that is 1.40 X 105 eV   which is .140 MeV

Part D)

The potential to create this energy is found by

qV = KE

V = (2.25 X 10-14)/(2)(1.6 X 10-19)

V = 7.03 X 104 V which can be rounded off to 70 kV

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