A cyclotron accelerates charged particles as shown in \({\rm{Figure 22}}{\rm{.64}}\). Using the results of the previous problem, calculate the frequency of the accelerating voltage needed for a proton in a\({\rm{1}}{\rm{.20 - T}}\) field.

The number of times a repeated event occurs per unit of time is referred to as frequency.

Apply the equation\({\rm{f = }}\frac{{{\rm{qB}}}}{{{\rm{2\pi m}}}}\)to find the frequency. Use the values,

\(\begin{array}{l}{\rm{m}} = {\rm{1}}{\rm{.67}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 27}}}}\;{\rm{kg}}\\{\rm{q}} = {\rm{1}}{\rm{.60}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 19}}}}\;{\rm{C}}\end{array}\)

\({\rm{f = }}\frac{{{\rm{qB}}}}{{{\rm{2\pi m}}}}\)

\({\rm{ = }}\frac{{\left( {{\rm{1}}{\rm{.60 \times 1}}{{\rm{0}}^{{\rm{ - 19}}}}\;{\rm{C}}} \right){\rm{ \times 1}}{\rm{.20}}\;{\rm{T}}}}{{{\rm{2\pi \times }}\left( {{\rm{1}}{\rm{.67 \times 1}}{{\rm{0}}^{{\rm{ - 27}}}}\;{\rm{kg}}} \right)}}\)

\({\rm{ = 1}}{\rm{.83 \times 1}}{{\rm{0}}^{\rm{7}}}\;{\rm{Hz}}\left( {\frac{{{\rm{MHz}}}}{{{\rm{1}}{{\rm{0}}^{\rm{6}}}\;{\rm{Hz}}}}} \right)\)

\({\rm{ = 18}}{\rm{.3}}\;{\rm{MHz}}\)

Therefore, frequency is \({\rm{18}}{\rm{.3}}\;{\rm{MHz}}\).