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

1μ= 1.661×10^−27kg

Suppose an alpha particle travels in a circular path of radius 4.50 cm in a uniform magnetic field with B = 1.20 T. Calculate
(a) its speed.
(b) its period of revolution.
(c) its kinetic energy.
(d) the potential difference through which it would have to be accelerated to achieve this energy.

Answer 1

Answer:

Explanation:

charge, q = 2e = 2 x 1.6 x 10^-19 C = 3.2 x 10^-19 C

mass, m = 4 u = 4 x 1.661 x 10^-27 kg = 6.644 x 10^-27 kg

Radius, r = 4.5 cm = 0.045 m

Magnetic field, B = 1.20 T

(a) Let the speed is v.

$v=\frac{Bqr}{m}$

$v=\frac{1.20\times 3.2\times 10^{-19}\times 0.045}{6.644\times 10^{-27}}$

v = 2.6 x 10^6 m/s

(b) Let T be the period of revolution

$T=\frac{2\pi r}{v}$

$T=\frac{2\times 3.14\times 0.045}{2.6\times 10^{6}}$

T = 1.09 x 10^-7 s

(c) The formula for the kinetic energy is

$K=\frac{B^{2}\times q^{2}\times r^{2}}{2m}$

$K=\frac{\left ( 1.20\times 3.2 \times 10^{-19}\times 0.045 \right )^{2}}{2\times 6.644\times 10^{-27}}$

K = 2.25 x 10^-14 J

(d) Let the potential difference is V.

K = qV

$V = \frac{K}{q}$

$V= \frac{2.25\times 10^-14}{3.2\times 10^{-19}}$

V = 70312.5 V