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?

Question

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Respuesta :

okpalawalter8 okpalawalter8 · Answer 1 · 2020-07-09T03:55:31+02:00

Answer:

The radius is [tex]r = 4.434 *10^{-5} \ m[/tex]

Explanation:

From the question we are told that

The magnetic field is [tex]B = 90 mT = 90*10^{-3} \ T[/tex]

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

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

[tex]\frac{mv^2}{r} = qvB[/tex]

=> [tex]r = \frac{m* v}{q * B}[/tex]

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

v is the escape velocity which is mathematically represented as

[tex]v = \sqrt{\frac{2 * KE}{m} }[/tex]

So

[tex]r = \frac{m}{qB} * \sqrt{\frac{2 * KE}{m} }[/tex]

apply indices

[tex]r = \frac{\sqrt{2 * KE * m} }{qB}[/tex]

substituting values

[tex]r = \frac{\sqrt{2 * 2.24*10^{-19}* 9.1 *10^{-31}} }{ 1.60 *10^{-19}* 90*10^{-3}}[/tex]

[tex]r = 4.434 *10^{-5} \ m[/tex]