Chapter 28, Problem 017 An alpha particle (q+2e, m-4.00 u) travels in a circular

Question

Chapter 28, Problem 017 An alpha particle (q+2e, m-4.00 u) travels in a circular path of radius 6.46 cm in a uniform magnetic field with B 1.77T. Calculate (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 (a) Number (b) Number (c) Number (d) Number Click if you would like to Show Work for this question: Open Show Work Units Units Units Units SHOW HINT LINK TO TEXT LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM VIDEO MINI-LECTURE

Explanation / Answer

a] The centripetal force should be provided by the magnetic force in order to achieve a stable orbit.

So, mv2/r = qvB

=> v = rqB/m

=> v = [0.0646 x 2 x 1.6 x 10-19 x 1.77]/(4 x 1.67 x 10-27) = 5.4775 x 106 m/s

b] v = 2 pi r/T

=> T = 2 x 3.14 x 0.0646 / 5.4775 x 106 = 7.41 x 10-8 s

c] Kinetic Energy = (1/2)mv2 = (1/2)(4 x 1.67 x 10-27)(5.4775 x 106 )2 = 1 x 10-13 J

d] Potential Energy needed to achieve this kinetic energy will thus be: W = qV = 1 x 10-13 J

and so the potential difference needed to accelerate the charges will be:

V = 1 x 10-13 / ( 2 x 1.6 x 10-19) = 313151.8 Volts.

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