Question

Cyclotrons are widely used in nuclear medicine for producing short-lived radioactive isotopes. These cyclotrons typically accelerate ${\mathbf{H}}^{\mathbf{-}}$(the hydride ion, which has one proton and two electrons) to an energy of $\mathbf{5}\mathbf{MeV}\mathbf{}\mathbf{to}\mathbf{}\mathbf{20}\mathbf{MeV}\mathbf{.}$This ion has a mass very close to that of a proton because the electron mass is negligible about $\frac{\mathbf{1}}{\mathbf{2000}}$of the proton’s mass. A typical magnetic field in such cyclotrons is $\mathbf{1}\mathbf{.}\mathbf{9}\mathbf{}\mathbf{T}\mathbf{.}$.(a) What is the speed of a $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{-}\mathbf{MeV}\mathbf{}{\mathbf{H}}^{\mathbf{-}}$? (b) If the ${\mathbf{H}}^{\mathbf{-}}$has energy $\mathbf{5}\mathbf{.0}\mathbf{MeV}\mathbf{}\mathbf{and}\mathbf{}\mathbf{B}\mathbf{=1.9}\mathbf{T}$ what is the radius of this ion’s circulator orbit?

Short answer

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

(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-

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

Determine the radius of this ion’s circulator orbit

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