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An alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 6.56 cm in a uniform magnetic field with B = 1.70 T. 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.

Respuesta :

Manetho

Explanation:

Data:

Charge of alpha particle, q = 2e

                                        = 2× 1.6 x 10^{-19 }

                                        = 3.2 x 10^{-19} C

Radius of circular path, r = 4.5 cm

                                     = 0.045 m

Magnetic field, B = 1.8 T

Mass of alpha particle, m = 4 u

                                       = 4×1.67 x 10^{-27 }

                                       = 6.68 x 10^{-27} kg

Solution:

(a)  Centripetal force = Force due to magnetic field

m v^2 / r = q v B

m v / r = q B

v = r q B / m

  = 0.045×3.2 x 10^{-19}×1.8 / 6.68 x 10^{-27 }

  = 3.88 x 10^6 m/s

Speed of alpha particle, v = 3.88 x 10^6 m/s

(b)

Time period, T = 2 π r / v

                       = 2× π × 0.045 / 3.88 x 10^6

                       = 7.2 x 10^{-8} s

Time period, T =7.2 x 10^-8 s

(c)   Kinetic energy, KE = (1/2) m v^2

                             = 0.5×6.68 x 10^{-27}×(3.88 x 10^6)^2

                             = 5.02 x 10^{-14}J

                             = ( 5.02 x 10^{-14} / 1.6 x 10^{-19} ) eV

                             = 3.14 x 10^5 eV

Kinetic energy, KE = 3.14 x 10^5 eV

(d)  Potential energy = Kinetic energy

                 q V = KE

             2 e  V = 3.14 x 10^5 e V

                  2V = 3.14 x 10^5 V

                     V = 1.57 x 10^5 V

Potential difference, V = 1.57 x 10^5 V

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