Question

Alpha particles, each having a charge of +2e and a mass of 6.64 ×10-27 kg, are accelerated in a uniform 0.50 T magnetic field to a final orbit radius of 0.50 m. The field is perpendicular to the velocity of the particles. What is the kinetic energy of an alpha particle in the final orbit? (1 eV = 1.60 × 10-19 J, e = 1.60 × 10-19 C)

Answer 1

Answer:

$KE=1.2036\times 10^{-12}\ J$

Explanation:

Given:

• charge on the alpha particle, $q=2e=3.2\times 10^{-19}\ C$

• mass of the alpha particle, $m=6.64\times 10^{-27}\ kg$

• strength of a uniform magnetic field, $B=0.5\ T$

• radius of the final orbit, $r=0.5\ m$

During the motion of a charge the magnetic force and the centripetal forces are balanced:

$q.v.B=m.\frac{v^2}{r}$

$m.v=q.B.r$

where:

v = velocity of the alpha particle

$v=\frac{q.B.r}{m}$

$v=\frac{3.2\times 10^{-19}\times 0.5\times 0.5}{6.64\times 10^{-27}}$

$v=1.2048\times 10^{7}\ m.s^{-1}$

Here we observe that the velocity of the aprticle is close to the velocity of light. So the kinetic energy will be relativistic.

We firstly find the relativistic mass as:

$m'=\frac{1}{\sqrt{1-\frac{v^2}{c^2} } } \times m$

$m'=\frac{6.64\times 10^{-27}}{\sqrt{1-\frac{(1.2048\times 10^7)^2}{(3\times 10^8)^2} } }$

$m'=6.6533\times10^{-27}\ kg$

now kinetic energy:

$KE=m'.c-m.c$

$KE=6.6533\times 10^{-27}\times (3\times 10^8)^2-6.64\times 10^{-27}\times (3\times 10^8)^2$

$KE=1.2036\times 10^{-12}\ J$