Question

A medical cyclotron used in the production of medical isotopes accelerates protons to 6.5 MeV. The magnetic field in the cyclotron is 1.2 T What is the diameter of the largest orbit, just before the protons exit the cyclotron?

Answer 1

Answer:

diameter of largest orbit is 0.60 m

Explanation:

given data

isotopes accelerates KE = 6.5 MeV

magnetic field B = 1.2 T

to find out

diameter

solution

first we find velocity from kinetic energy equation

KE = 1/2 × m×v²   ........1

6.5 × 1.6 × $10^{-19}$ = 1/2 × 1.672 × $10^{-27}$ ×v²

v = 3.5 × $10^{7}$ m/s

so

radius will be

radius = $\frac{m*v}{B*q}$   ........2

radius =  $\frac{1.672*10^{-27}*3.5*10^{7}}{1.2*1.6*10^{-19}}$  

radius = 0.30

so diameter = 2 × 0.30

so diameter of largest orbit is 0.60 m

Answer 2

Answer:

The diameter of the largest orbit is 6.52m

Explanation:

First, we will use only magnitudes, as the problem is not asking for directions or any vectorial result.

Second, besides the given data (K=6.5 MeV, B=1.2 T), we have to know that the mass of a proton is 938.3 MeV/C^2, or 1.6726 x 10-27 kg, and that the charge of an electron in magnitude is q=1.6 x 10-19 C.

With this in mind, we use the Lorent'z force:

$F=qvB$

where q is the electron's charge, v is the velocity of the particles and B is the magnetic field. We also use the definition of force in a circular movement:

$F=mv^2r$

where m is the mass of the particle and r is the radius. Then, by equalizing both expressions we can clear the radius r

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

and as we can see, we have that data except for the velocity, wich we can calculate using that

$K=\frac{1}{2}mv^2\Leftrightarrow v=\sqrt{\frac{2K}{m}}$

then

$v=\sqrt{\frac{2*6.5Mev}{938.3\frac{Mev}{c^2}}} =35287551\frac{m}{s}$

which is the velocity of the protons, and which we replace in our previous result

$r=\frac{1.6*10^{-19}*1.2}{1.6727*10^{-27}*35287551}=3.26m$

Finally, to calculate the diameter, we multiply r times 2.