Chapter 4: Q20E (page 913)

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 $\frac{1}{2000}$ 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 circulator orbit?

Short Answer

Expert verified

(a) The speed of a $5.0_MeV H^{-} is 3.1 \times 10^{7} m / s$

(b) The radius of this ion’s circulator orbit is 0.17 m

Step by step solution

Definition of isotopes

The term isotopes may be defined as the allotropes having same atomic number but different mass number.

Determine the speed of a 5.0-MeV H-

$According to law of conservation of {energyK}_{i} + U_{i} = K_{f} + U_{f}, K_{i} = 0 so K_{f} = U_{f} - U_{i} put given value K_{f} = 5.0 MeV Now K_{f} = 5.0 \times 1.6 \times 10^{- 19} \times 10^{6} J Further \frac{1}{2} {mv}^{2} = 5.0 \times 1.6 \times 10^{- 19} \times 10^{6} J v = \sqrt{\frac{2}{m} \times 1.6 \times 10^{- 19} \times 10^{6} J} v = 3.1 \times 10^{7} m / s Hence, the speed of a 5.0 - MeV H^{-} is 3.1 \times 10^{7} m / s$

Determine the radius of this ion’s circulator orbit

$For the radius using relation R = \frac{(1.67 \times 10^{- 27} kg) (3.1 \times 10^{7} m / s)}{(1.6 \times 10^{- 19}) (1.9 T)} R = 0.17 m Hence, the radius of this ion ’ s circulator orbit is 0.17 m$