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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?

Respuesta :

Answer:

[tex]r=5.278\times 10^{-4}\ m[/tex]

Explanation:

Given that:

  • magnetic field intensity, [tex]B=0.07\ T[/tex]
  • kinetic energy of electron, [tex]KE=1.2\ eV= 1.2\times 1.6\times 10^{-19}\ J= 1.92\times 10^{-19}\ J[/tex]
  • we have mass of electron, [tex]m=9.1\times 10^{-31}\ kg[/tex]

Now, form the mathematical expression of Kinetic Energy:

[tex]KE= \frac{1}{2} m.v^2[/tex]

[tex]1.92\times 10^{-19}=0.5\times 9.1\times 10^{-31}\times v^2[/tex]

[tex]v^2=4.2198\times 10^{11}[/tex]

[tex]v=6.496\times 10^6\ m.s^{-1}[/tex]

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

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

[tex]r=\frac{9.1\times 10^{-31}\times 6.496\times 10^6 }{1.6\times 10^{-19}\times 0.07}[/tex]

[tex]r=5.278\times 10^{-4}\ m[/tex]

onyebuchinnaji

The radius of orbit of the cyclotrons in the given field is 5.287 x 10⁻⁵ m.

Force of the electron

The magnetic force and centripetal force of the electron is calculated as follows;

qvB = mv²/r

qB = mv/r

r = mv/qB

where;

  • m is mass of electron
  • v is speed of the electron
  • q is the magnitude of the charge
  • B is magnetic field strength

Kinetic energy of the electron

The kinetic energy of the electron is given as;

K.E = ¹/₂mv²

mv² = 2K.E

v² = 2K.E/m

v² = (2 x 1.2 x 1.6 x 10⁻¹⁹)/(9.11 x 10⁻³¹)

v² = 4.215 x 10¹¹

v = √4.215 x 10¹¹

v = 6.5 x 10⁵ m/s

Radius of their orbit

The radius of their orbit is calculated as follows;

r = mv/qB

r = (9.11 x 10⁻³¹ x 6.5 x 10⁵) / (1.6 x 10⁻¹⁹ x 70 x 10⁻³)

r = 5.287 x 10⁻⁵ m

Thus, the radius of orbit of the cyclotrons in the given field is 5.287 x 10⁻⁵ m.

Learn more about magnetic force here: https://brainly.com/question/13277365

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