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

Part a)

`v = 3.16 * 10^7 m/s`

Part b)

r = 0.166 m

Step-by-step explanation:

Part a)

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

`E = 5 MeV`

here we have

`(1)/(2)mv^2 = 5* 10^6(1.6 * 10^(-19))`

also we know that

`m = 1.6 * 10^(-27) kg`

now we have

`v = \sqrt{(2 * 5 * 10^6(1.6 * 10^(-19))/(1.6* 10^(-27))`

`v = 3.16 * 10^7 m/s`

Part b)

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

so we have

`qvB = (mv^2)/(r)`

so we have

`r = (mv)/(qB)`

so we have

`r = (1.6 * 10^(-27)(3.16 * 10^7))/((1.6 * 10^(-19)) 1.9)`

`r = 0.166 m`