Question

A medical cyclotron used in the production of medical isotopes accelerates protons to 6.5 MeV. The magnetic field in the cyclotron is 1.9 T.
(a) What is the diameter of the largest orbit, just before the protons exit the cyclotron? Express your answer with the appropriate units. d = 57 cm Previous
(b) A proton exits the cyclotron 1.0 ms after starting its spiral trajectory in the center of the cyclotron. How many orbits does the proton complete during this 1.0 ms?

Answer 1

Given Information"

Kinectic energy = 6.5 MeV

magnetic field = B = 1.9 T

Period = T = 1 ms

Required Information"

Diameter = d = ?

Answer:

d = 38.6 cm

f = 1000

Explanation:

(a) What is the diameter of the largest orbit, just before the protons exit the cyclotron?

The radius and hence the diameter of the largest orbit can be found by using

r = mv /|q|B

Where q = 1.60x10⁻¹⁹ C and m = 1.67x10⁻²⁷ kg and v is the speed of the particle  which can be found by using

v = √2K /m

Where K is the kinetic energy in Joules.

1 eV is equal to 1.60x10⁻¹⁹ J

K = 6.5x10⁶*1.60x10⁻¹⁹ = 1.04x10⁻¹² J

v = √2*1.04x10⁻¹² /1.67x10⁻²⁷

v = 35.29x10⁶ m/s

r = 1.67x10⁻²⁷*35.29x10⁶ /1.60x10⁻¹⁹*1.9

r = 0.193 m

d = 2r

d = 2*0.193 = 0.386

or d = 38.6 cm

(b) A proton exits the cyclotron 1.0 ms after starting its spiral trajectory in the center of the cyclotron. How many orbits does the proton complete during this 1.0 ms?

The period T and frequency (number of rotations per second) are related as

f = 1/T

f = 1/0.001

f = 1000

Answer 2

The diameter of the largest orbit just before the protons exit the cyclotron is 39 cm. The number of orbits completed by the proton during this 1.0 ms is 14000 revolutions.

Given that:

• the kinetic energy of the medical isotopes = 6.5 MeV

The kinetic energy for the protons can be computed by using the formula:

$\mathbf{K.E = \dfrac{1}{2} mv^2}$

mv² = 2 K.E

$\mathbf{v^2 = \dfrac{2K.E}{m}}$

$\mathbf{v =\sqrt{ \dfrac{2K.E}{m}}}$

$\mathbf{v =\sqrt{ \dfrac{2\times (6.5 \ MeV)}{1.672 \times 10^{-27} \ kg }}}$

$\mathbf{v =\sqrt{ \dfrac{2\times (6.5 \times 1.6 \times 10^{-13} \ J )}{1.672 \times 10^{-27} \ kg }}}$

$\mathbf{v = \sqrt{1.24401914\times 10^{15}}}$

v = 3.53 × 10⁷ m/s

However, the radius of the orbit can be estimated by using the formula:

$\mathbf{\dfrac{mv^2}{r} = qvB}$

Making radius (r) the subject, we have:

$\mathbf{r = \dfrac{mv}{qB}}$

$\mathbf{r = \dfrac{(1.672 \times 10^{-27}) \times (3.53 \times 10^7\ m/s)}{(1.9 T) (1.6 \times 10^{-19\ C})}}$

$\mathbf{r = \dfrac{(5.90216\times 10^{-20})}{(3.04 \times 10^{-19\ C})}}$

r = 0.19415 m

Since diameter (D) = 2r, then the diameter of the largest orbit is:

= 2(0.19415  m)

= 0.39 m

≅ 39 cm

The time period to complete a revolution around the spiral trajectory is:

$\mathbf{T = \dfrac{2 \pi r}{v}}$

$\mathbf{T = \dfrac{2 \pi\times 0.39}{3.53 \times 10^7}}$

T = 0.7 × 10⁻⁷ s

Finally, the number of orbits that the proton does to complete the revolution in 1 ms is:

$\mathbf{n = \dfrac{t}{T}}$

$\mathbf{n = \dfrac{1 \times 10^{-3}}{0.7 \times 10^{-7}}}$

n = 14285.71

n ≅ 14000 revolutions

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