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 ionize it. One such source uses a magnetic field of 90 mTmT, and the electrons' kinetic energy is 1.4 eV.
Required:
If the electrons travel in a plane perpendicular to the field, what is the radius of their orbits?

Answer 1

Answer:

The radius is $r = 4.434 *10^{-5} \ m$

Explanation:

From the question we are told that

    The magnetic field is  $B = 90 mT = 90*10^{-3} \ T$

     The electron kinetic energy is  $KE = 1.4 eV = 1.4 * (1.60*10^{-19}) =2.24*10^{-19} \ J$

Generally for the collision to occur the centripetal force of the electron in it orbit is equal to the magnetic force applied  

   This is mathematically represented as

   $\frac{mv^2}{r} = qvB$

=>    $r = \frac{m* v}{q * B}$

Where  m is the mass of electron with values $m = 9.1 *10^{-31} \ kg$  

             v is the escape velocity  which is mathematically represented as

                $v = \sqrt{\frac{2 * KE}{m} }$

So  

       $r = \frac{m}{qB} * \sqrt{\frac{2 * KE}{m} }$

     apply indices

    $r = \frac{\sqrt{2 * KE * m} }{qB}$

substituting values

   

        $r = \frac{\sqrt{2 * 2.24*10^{-19}* 9.1 *10^{-31}} }{ 1.60 *10^{-19}* 90*10^{-3}}$

       $r = 4.434 *10^{-5} \ m$