Irodov 3.398: Counting the Turns of a Cyclotron – Amit & Deepali Sinha Foundation

Problem Irodov 3.398

Singly charged helium ions He $^+$ are accelerated in a cyclotron up to the largest orbit radius $r = 60$ cm. The oscillator frequency is $\nu = 10.0$ MHz and the effective accelerating voltage across the dees is $V = 50$ kV. Neglecting the width of the gap between the dees, find:

the total time the ion spends being accelerated;
the approximate length of the path it covers during the acceleration.

The cyclotron: two dees across a narrow gap, fed by the oscillator, in the uniform field B; r is the largest orbit radius of the ions. — The cyclotron: two dees across a narrow gap, fed by the oscillator, in the uniform field $\vect B$ ; $r$ is the largest orbit radius of the ions.

Given

He $^+$ , charge $e$
$r = 60$ cm
$\nu = 10.0$ MHz
$V = 50$ kV
gap neglected

Find

(a) the time of acceleration

t

· (b) the path length

s

Solution

Step 1The final energy

The ion revolves at the oscillator frequency $\nu$ on every orbit, so on the largest one its speed is the circumference times $\nu$ :

v_{\mathrm{max}} = 2\pi\nu r, \qquad T = \frac{m v_{\mathrm{max}}^2}{2} = 2\pi^2 m\nu^2 r^2 \tag{1}

Step 2The number of revolutions

In each revolution the ion crosses the gap twice and gains $eV$ each time, so its energy grows by $2eV$ per revolution. Dividing the final energy (1) by that gain:

N = \frac{T}{2eV} = \frac{\pi^2 m\nu^2 r^2}{eV} \tag{2}

Step 3(a) The time of acceleration

Every revolution takes the same time $1/\nu$ , and with the gap neglected nothing else takes time, so $t = N/\nu$ with $N$ from (2). Numerically, with $m = 6.64\times10^{-27}$ kg (four atomic mass units) and $e = 1.60\times10^{-19}$ C, $N = 295$ revolutions and

t = \frac{\pi^2(6.64\times10^{-27})(10.0\times10^{6})(0.60)^2}{(1.60\times10^{-19})(50\times10^{3})}\ \text{s}

t = \frac{\pi^2 m\nu r^2}{eV} \approx 30\ \mu\text{s}

(3)

Step 4(b) The path covered

The ion's path: a semicircle in each dee, the radius growing at every gap crossing out to the largest one, r. — The ion’s path: a semicircle in each dee, the radius growing at every gap crossing out to the largest one, $r$ .

Number the gap crossings $k = 1, \ldots, n$ , with $n = 2N$ in all. After the $k$ th crossing the energy is $keV$ , so the speed is $v_{\mathrm{max}}\sqrt{k/n}$ , and the following half-turn, lasting $1/(2\nu)$ , covers $v_k/(2\nu)$ . With $n$ large the sum of square roots becomes an integral:

s = \sum_{k=1}^{n} \frac{v_{\mathrm{max}}}{2\nu}\sqrt{\frac{k}{n}} \approx \frac{v_{\mathrm{max}}}{2\nu\sqrt{n}}\int_0^{n}\sqrt{k}\,dk = \frac{v_{\mathrm{max}}}{2\nu\sqrt{n}}\cdot\frac{2}{3}\,n^{3/2} = \frac{2}{3}\,v_{\mathrm{max}}\,\frac{N}{\nu}

So $s$ is two thirds of $v_{\mathrm{max}} t$ . Using $v_{\mathrm{max}}$ from (1) and $t$ from (3):

s = \frac{2}{3}\,(2\pi\nu r)\,\frac{\pi^2 m\nu r^2}{eV}

Numerically, with $v_{\mathrm{max}} = 3.77\times10^{7}$ m/s and $t = 29.5\ \mu$ s,

s = \frac{2}{3}\,(3.77\times10^{7})(29.5\times10^{-6})\ \text{m}

s \approx \frac{4\pi^3 m\nu^2 r^3}{3eV} \approx 0.74 \text{ km}

(4)

Check

Energy: (1) gives $T = 4.72\times10^{-12}$ J $= 29.5$ MeV, and $295$ revolutions at $2eV = 0.10$ MeV each make the same $29.5$ MeV. The speed grows as the square root of the time (the energy grows linearly in time), and the mean of $\sqrt{\tau}$ over $0 < \tau < 1$ is $2/3$ , which is the factor in $s = (2/3)\,v_{\mathrm{max}} t$ read from the other side. The non-relativistic treatment holds: $29.5$ MeV is under 1 per cent of the ion’s rest energy, about $3.7$ GeV, and $v_{\mathrm{max}}/c = 0.13$ .

Remark

In thirty microseconds the ion runs about three quarters of a kilometre inside a machine little more than a metre across. That is why a cyclotron needs a high vacuum: the residual gas must leave the ions a mean free path much longer than this spiral, or they are scattered out of step long before they reach the rim.