Question

In a cyclotron, the orbital radius of protons with energy 300KeV is 16.0cm . You are redesigning the cyclotron to be used instead for alpha particles with energy 300 KeV . An alpha particle has charge q = +2e and mass m =$\mathbf{6}\mathbf{.}\mathbf{64}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{27}}\mathbf{kg}$ . If the magnetic field isn’t changed, what will be the orbital radius of the alpha particles?

Short answer

The orbital radius of the alpha particles is 16.0 cm

Step-by-step solution

Definition of Magnetic field

The term magnetic field may be defined as the area around the magnet behave like a magnet.

Determine the orbital radius of the alpha particles

The radius formula used $\mathrm{R}=\frac{\sqrt{2\mathrm{mK}}}{\left|\mathrm{q}\right|\mathrm{B}}$

For proton${\mathrm{R}}_{\mathrm{P}}=\frac{\sqrt{2{\mathrm{m}}_{\mathrm{p}}\mathrm{K}}}{\mathrm{eB}}$and for alpha particle ${\mathrm{R}}_{\mathrm{\alpha }}=\frac{\sqrt{2{\mathrm{m}}_{\mathrm{\alpha }}\mathrm{K}}}{\mathrm{eB}}$

Divide ${\mathrm{R}}_{\mathrm{\alpha }}$by ${\mathrm{R}}_{\mathrm{p}}$get

$$\begin{gathered} \frac{{\mathrm{R}}_{\mathrm{\alpha }}}{{\mathrm{R}}_{\mathrm{p}}}=\frac{2\mathrm{eB}\sqrt{2{\mathrm{m}}_{\mathrm{\alpha }}\mathrm{K}}}{2\mathrm{eB}\sqrt{2{\mathrm{m}}_{\mathrm{p}}\mathrm{K}}} \\ \frac{{\mathrm{R}}_{\mathrm{\alpha }}}{{\mathrm{R}}_{\mathrm{p}}}=\frac{1\sqrt{{\mathrm{m}}_{\mathrm{\alpha }}}}{2\sqrt{{\mathrm{m}}_{\mathrm{p}}}} \end{gathered}$$

The mass of proton is $\frac{1}{4}$time mass of the alpha particle so

$$\begin{gathered} \frac{{\mathrm{R}}_{\mathrm{\alpha }}}{{\mathrm{R}}_{\mathrm{p}}}=\frac{1\sqrt{4{\mathrm{m}}_{\mathrm{p}}}}{2\sqrt{{\mathrm{m}}_{\mathrm{p}}}} \\ {\mathrm{R}}_{\mathrm{p}}=16.0\mathrm{cm} \\ {\mathrm{R}}_{\mathrm{\alpha }}=16.0\mathrm{cm} \end{gathered}$$

Hence, the orbital radius of the alpha particles is 16.0 cm .