Question

Cyclotrons are widely used in nuclear medicine for producing short-lived radioactive isotopes. These cyclotrons typically accelerate H− (the hydride ion, which has one proton and two electrons) to an energy of 5MeV to 20MeV. This ion has a mass very close to that of a proton because the electron mass is negligible−about 1/2000 of the protons mass. A typical magnetic field in such cyclotrons is 1.7 T .
Part A

What is the speed of a 5.0-MeV H−?

Express your answer with the appropriate units.

Part B

If the H− has energy 5.0MeV and B= 1.7 T , what is the radius of this ions circular orbit?

Express your answer with the appropriate units.

Answer 1

Answer:

(A) $v=3.1\times 10^7 m.s^{-1}$

(B) $r=0.1903 m$

Explanation:

Given:

• initial energy of hydride ion, $KE_i=5 MeV$

• final energy of hydride ion, $KE_i=20 MeV$

• mass of ion (=mass of proton, we know), $m=1.67\times 10^{-27} kg$

• magnetic field, $B=1.7 T$

(A)

kinetic energy of 5 MeV ion:

$KE_i=5 MeV$

$\frac{1}{2} m.v^2=5\times 10^{6}\times 1.6\times 10^{-19}$

$\frac{1}{2} 1.67\times 10^{-27}.v^2=5\times 10^{6}\times 1.6\times 10^{-19}$

$v=3.1\times 10^7 m.s^{-1}$

(B)

Radius of ion when having energy 5.0 MeV:

We know:

$F=q.v.B$..................................(1)

also,

$F=m.\frac{v^2}{r}$..........................(2)

where:

F= force on the charge

q= quantity of charge on the particle (an electron in this case)

m= mass of the charged particle

B= magnetic field in which the charge is projected

r= radius of the path circulation of the charge

v= velocity of the charge at the instant

From eq. (1) & (2)

$m.\frac{v^2}{r}=q.v.B$

$r=\frac{m.v}{q.B}$

putting the reaspective values:

$r=\frac{1.67\times 10^{-27}\times 3.1\times 10^7}{1.6\times 10^{-19}\times 1.7}$

$r=0.1903 m$