An alpha particle ( q = +2 e , m =4.00 u) travels in a circular path of radius 5

Question

An alpha particle (q = +2e, m =4.00 u) travels in a circular path of radius 5.61 cm in a uniformmagnetic field with B = 1.27 T. Calculate(a) its speed, (b) its period ofrevolution, (c) its kinetic energy (in eV), and(d) the potential difference (in V) through whichit would have to be accelerated to achieve this energy
An alpha particle (q = +2e, m =4.00 u) travels in a circular path of radius 5.61 cm in a uniformmagnetic field with B = 1.27 T. Calculate(a) its speed, (b) its period ofrevolution, (c) its kinetic energy (in eV), and(d) the potential difference (in V) through whichit would have to be accelerated to achieve this energy

Explanation / Answer

The charge on the alpha particle is q = +2e = +2 * 1.6 *10-19 C = 3.2 * 10-19 C The mass of the alpha particle is m = 4.00 u = 4.00 * 1.66 *10-27 kg = 6.64 * 10-27 kg The alpha particle travels in a circular path of radius r =5.61 cm = 5.61 * 10-2 m The alpha particle travels in a uniform magnetic field with B= 1.27 T (a)The centripetal force on the alpha particle is F = (mv2/r) ----------------(1) The force on the alpha particle due to the magnetic fieldis FB= q * v * B ----------------(2) From equations (1) and (2),we get (mv2/r) = q * v * B or v = (q * B * r/m) Substituting the values in the above equation,we get v = (3.2 * 10-19 * 1.27 * 5.61 *10-2/6.64 * 10-27) or v = 3.43 * 106 m/s Therefore,the speed of the alpha particle in the circular pathis v = 3.43 * 106 m/s. (b)The angular frequency of the alpha particle is w = (B * q/m) or (2/T) = (B * q/m) or T = (2m/B * q) Susbtituting the values in the above equation,we get T = (2 * 3.14 * 6.64 * 10-27/1.27 * 3.2 *10-19) or T = 10.26 * 10-8 s = 102.6 * 10-9 s =102.6 ns Therefore,the period of its revolution is T = 102.6 ns (c)The kinetic energy of the alpha particle is K = (1/2)mv2 or K = (1/2) * 6.64 * 10-27 * (3.43 *106)2 or K = 39.06 * 10-15 J We know that 1eV = 1.6 * 10-19 J or 1J = (1/1.6 * 10-19) eV Therefore,we get K = 39.06 * 10-15 * (1/1.6 * 10-19)eV or K = 244125 eV (d)Let the potential difference through which the alphaparticle would have to be accelerated to achieve this energy beVo,therefore,we get K = e * Vo or (1/2)mv2 = e * Vo or Vo= (1/e) * (1/2)mv2 or Vo= (39.06 * 10-15/1.6 *10-19) or Vo= 244125 V or Vo= (1/e) * (1/2)mv2 or Vo= (39.06 * 10-15/1.6 *10-19) or Vo= 244125 V

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.