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kozerog [31]
2 years ago
11

A proton exits the cyclotron 1.0 ms after starting its spiral trajectory in the center of the cyclotron. How many orbits does th

e proton complete during this 1.0 ms?
Physics
1 answer:
NeX [460]2 years ago
5 0

(a) The diameter of the largest orbit just before the protons exit the cyclotron is 39 cm.

(b) The number of orbits completed by the proton during this 1.0 ms is 14000 revolutions.

The energy that an object has as a result of motion is known as kinetic energy. It is described as the effort required to move a mass-determined body from rest to the indicated velocity.

K.E = 1/2mv²

mv² = 2 K.E

v = sqrt( 2*KE / M)

(a) The KE of the medical isotopes = 6.5 MeV

v = sqrt( 2* 6.5* 1.6* 10^-13 / 1.67* 10^-27 )

v = 3.53 × 10⁷ m/s

Now the centripetal force:

mv² / R = qvB

r = q*v / mv²

r = ( 1.67* 10^-27 * 3.53 × 10⁷ ) / ( 1.9* 1.6* 10^-19 )

r = 0.19415 m

diameter d = 2r,

d= 2(0.19415  m)

d= 0.39 m ≅ 39 cm

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

T = 2πr / v

T = 2*3.14* 0.1941 / 3.53*10^7

T = 0.7 × 10⁻⁷ s

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

n = t / T

n = 10^-3 / (0.7*10^-7)

n = 14285.71  ≅ 14000 revolutions

The complete question is :

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.

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

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?

To know more about cyclotron refer to:  brainly.com/question/14985809

#SPJ4

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Answer:

The distance is

=

7

m

Explanation:

Apply the equation of motion

s

(

t

)

=

u

t

+

1

2

a

t

2

The initial velocity is

u

=

0

m

s

−

1

The acceleration is

a

=

2

m

s

−

2

Therefore, when

t

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3

s

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s

(

3

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=

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+

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2

⋅

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and when

t

=

4

s

s

(

4

)

=

0

+

1

2

⋅

2

⋅

4

2

=

16

m

Therefore,

The distance travelled in the fourth second is

d

=

s

(

4

)

−

s

(

3

)

=

16

−

9

=

7

m

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