An alpha particle ( q = +2 e , m = 4.00 u)travels in a circular path of radius 7

Question

An alpha particle (q = +2e, m = 4.00 u)travels in a circular path of radius 7.00cm in a magnetic field with B = 1.60 T. Calculate the following values. (a) the speed of the particle
1 m/s

(b) its period of revolution
2 s

(c) its kinetic energy
3 eV

(d) the potential difference through which it would have to beaccelerated to achieve this energy
4 V (a) the speed of the particle
1 m/s

(b) its period of revolution
2 s

(c) its kinetic energy
3 eV

(d) the potential difference through which it would have to beaccelerated to achieve this energy
4 V

Explanation / Answer

For an - particle : .    Charge is : (q) = 3.2 * 10-19  C ; mass (m) = 6.68 * 10-27   kg .    Magnetic field (B) = 1.60 T ; radius (r) = 0.07 m . a) Speed of the particle is :    m v2 / r   = q vB       or       v = q B r / m = 3.2 x 10-19 * 1.60 * 0.07 / 6.68 x10-27             = 5.365 x 106 m/s (b) distance per revolution = 2r = 2 * 0.07 = 0.4396 meters      period = distance moved /speed = 0.4396 / 5.365 x 106 = 0.0819 x 10-6 seconds (c) KE = (1/2) m v2 = (1/2) * 6.68 x10-27 * (5.365 x 106 )2    = 9.613 x10-14 Joules (d)    potential difference?    KE = qV     V = KE / q = 9.613 x10-14 / 3.2 x 10-19 = 3.00x 105  volts . Hope this helps u!    = 9.613 x10-14 Joules (d)    potential difference?    KE = qV     V = KE / q = 9.613 x10-14 / 3.2 x 10-19 = 3.00x 105  volts . Hope this helps u!

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