Question

Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e., protons with one orbiting electron) are fed into a chamber where there is a strong magnetic field. Electrons in this chamber are trapped in tight orbits, which greatly increases the chance that they will collide with a hydrogen atom and ionizes it. One such source uses a magnetic field of 70 mT, and the electrons' kinetic energy is 1.2 eV. If the electrons travel in a plane perpendicular to the field, what is the radius of their orbits?

Answer 1

Answer:

$r=5.278\times 10^{-4}\ m$

Explanation:

Given that:

• magnetic field intensity, $B=0.07\ T$

• kinetic energy of electron, $KE=1.2\ eV= 1.2\times 1.6\times 10^{-19}\ J= 1.92\times 10^{-19}\ J$

• we have mass of electron, $m=9.1\times 10^{-31}\ kg$

Now, form the mathematical expression of Kinetic Energy:

$KE= \frac{1}{2} m.v^2$

$1.92\times 10^{-19}=0.5\times 9.1\times 10^{-31}\times v^2$

$v^2=4.2198\times 10^{11}$

$v=6.496\times 10^6\ m.s^{-1}$

from the relation of magnetic and centripetal forces we have the radius as:

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

$r=\frac{9.1\times 10^{-31}\times 6.496\times 10^6 }{1.6\times 10^{-19}\times 0.07}$

$r=5.278\times 10^{-4}\ m$