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

A cyclotron accelerates protons. Its two hollow dees lie in a uniform magnetic field of $1.20\ \text{T}$ , at right angles to the dees, and the radius of the dees is $50\ \text{cm}$ , as in the figure. An oscillator applies an alternating voltage of peak value $20\ \text{kV}$ across the narrow gap between the dees. A proton starts almost at rest near the centre. Each time it crosses the gap, the voltage across the gap is at its peak and in the right direction to speed it up. The gap is so narrow that the time spent crossing it can be ignored. Take $m_p = 1.67 \times 10^{-27}\ \text{kg}$ , $e = 1.60 \times 10^{-19}\ \text{C}$ and $1\ \text{MeV} = 1.60 \times 10^{-13}\ \text{J}$ , and ignore the change of mass with speed.

Top view of a cyclotron. Two hollow D-shaped dees face each other across a narrow gap, together making a circle of radius 50 cm. Small crosses all over the picture show a uniform magnetic field of 1.20 T going into the page. A source sits at the centre, and a sketch of a path of growing half circles starts there. Wires join the two dees to an oscillator below them, marked 20 kV peak.

At what frequency must the oscillator run?
What is the kinetic energy of the protons as they leave the dees?
How many complete turns does a proton make inside the dees before it leaves?
Each turn takes one period of the oscillator. How long do those complete turns take?
The same machine, with the same magnetic field, is now used for alpha particles (charge $+2e$ , mass $6.64 \times 10^{-27}\ \text{kg}$ ). At what frequency must the oscillator now run?
What is the kinetic energy of the alpha particles as they leave?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. In the field the proton moves in circles. Their period does not depend on the speed, so the oscillator runs at the cyclotron frequency. The largest radius fixes the final energy, and the energy gained per turn counts the turns.

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

\displaystyle K = \frac{(eBR)^2}{2m_p} = \frac{(9.60 \times 10^{-20}\ \text{kg m/s})^2}{2 \times 1.67 \times 10^{-27}\ \text{kg}} = 2.76 \times 10^{-12}\ \text{J} = 17.2\ \text{MeV}

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 $q^2/m$ :

\displaystyle K_\alpha = \frac{(2eBR)^2}{2m_\alpha} = 17.3\ \text{MeV}

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

Answer: (a) $18.3\ \text{MHz}$ . (b) $17.2\ \text{MeV}$ . (c) $431$ complete turns. (d) $23.6\ \mu\text{s}$ . (e) $9.20\ \text{MHz}$ . (f) $17.3\ \text{MeV}$ .

Watch out. A proton gains $eV$ at each crossing, not each turn: counting one crossing per turn gives twice as many turns. And doubling the charge does not double the energy, $34.5\ \text{MeV}$ : the alpha particle is also about four times as heavy.

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