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7 days ago

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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 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?

1 answer:

Keith_Richards [2.9K]7 days ago

6 0

Answer:

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

Explanation:

The problem states that

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

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

In general, for a collision to happen, the centripetal force on the electron in its orbit must equal the magnetic force acting on it

This can be mathematically expressed as

$\frac{mv^2}{r} = qvB$

=> $r = \frac{m* v}{q * B}$

Where m denotes the electron’s mass, which has a value of $m = 9.1 *10^{-31} \ kg$

v signifies the escape velocity, mathematically represented as

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

Thus,

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

applying indices

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

substituting these values

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

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

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