A cyclotron accelerates protons. Its two hollow dees lie in a uniform magnetic field of 1.20 T, at…

Step 1. Frequency (a). $f = \dfrac{eB}{2\pi m_p} = \dfrac{1.60 \times 10^{-19}\ \text{C} \times 1.20\ \text{T}}{2\pi \times 1.67 \times 10^{-27}\ \text{kg}} = 18.3\ \text{MHz}$.

Step 2. Final energy (b). At the edge $r = R$, so $m_pv = eBR$ and

The final speed is $5.75 \times 10^{7}\ \text{m/s}$, about a fifth of the speed of light. The mass would grow by only about $2\%$, so leaving relativity out is fair here.

Step 3. Turns (c). Each turn has two crossings of the gap, so it adds $2eV = 40\ \text{keV}$. The number of turns is $\dfrac{17246\ \text{keV}}{40\ \text{keV}} = 431.1$, so the proton makes $431$ complete turns before it leaves.

Step 4. Time (d). Every turn takes $1/f$, whatever its radius: $t = \dfrac{431}{18.3 \times 10^{6}\ \text{Hz}} = 23.6\ \mu\text{s}$. The last part turn adds less than $0.1\ \mu\text{s}$.

Step 5. Alpha particles (e, f). The frequency is $\dfrac{qB}{2\pi m}$ with $q = 2e$ and $m = 6.64 \times 10^{-27}\ \text{kg}$: $f_\alpha = 9.20\ \text{MHz}$, about half the proton's. The final energy is $\dfrac{(qBR)^2}{2m}$, which grows as $q^2/m$:

almost the same as for the proton, because $q^2/m$ is almost the same.