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 = 1.661 ✕ 10⁻²⁷ kg. Suppose an alpha particle travels in a circular path of radius 4.60 cm in a uniform magnetic field with B= 1.70 T. Calculate the following values.

(a) its speed
  m/s

(b) its period of revolution
s

(c) its kinetic energy
eV

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

Answer 1

a)

r = 0.046 m

B = 1.70 T

m = mass = 4 u = 4 x 1.661 x 10^-27 kg

q = charge = 2 e = 2 x 1.6 x 10^-19

v = speed

using the formula

r = mv /qB

v = qBr/m

v = (2 x 1.6 x 10^-19) (1.70) (0.046) /(4 x 1.661 x 10^-27)

v = 3.8 x 10⁶ m/s

b)

period of revolution is given as

T = 2 $\pi$ m/(qB)

T = 2 (3.14) (4 x 1.661 x 10^-27 )/((2 x 1.6 x 10^-19) (1.70)) = 7.7 x 10^-8 sec

c)

kinetic energy is given as

KE = (0.5) m v² = (0.5) (4 x 1.661 x 10^-27 ) (3.8 x 10⁶ )² = 4.8 x 10^-14 J = 4.8 x 10^-14/(1.6 x 10^-19) = 3 x 10⁵ eV

d)

using conservation of energy

electric potential energy = kinetic energy

q $\Delta$ V = KE

(2 x 1.6 x 10^-19) $\Delta$ V = 4.8 x 10^-14

$\Delta$ V = 1.5 x 10⁵ volts