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 5 MeV to 20 MeV. This ion has a mass very close to that of a proton because the electron mass is negligible—about 1 2000 of the proton’s mass. A typical magnetic field in such cyclotrons is 1.9 T. (a) What is the speed of a 5.0-MeV H-? (b) If the H- has energy 5.0 MeV and B = 1.9 T, what is the radius of this ion’s circular orbit?

Answer 1

Answer:

Part a)

$v = 3.16 \times 10^7 m/s$

Part b)

r = 0.166 m

Explanation:

Part a)

As we know that the energy of the Hydride ion is given as

$E = 5 MeV$

here we have

$\frac{1}{2}mv^2 = 5\times 10^6(1.6 \times 10^{-19})$

also we know that

$m = 1.6 \times 10^{-27} kg$

now we have

$v = \sqrt{\frac{2 \times 5 \times 10^6(1.6 \times 10^{-19}}{1.6\times 10^{-27}}$

$v = 3.16 \times 10^7 m/s$

Part b)

As we know that magnetic force on the charge is centripetal force

so we have

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

so we have

$r = \frac{mv}{qB}$

so we have

$r = \frac{1.6 \times 10^{-27}(3.16 \times 10^7)}{(1.6 \times 10^{-19}) 1.9}$

$r = 0.166 m$