Irodov 3.399: One Rest Energy per Gap Crossing – Amit & Deepali Sinha Foundation

Problem Irodov 3.399

The period of revolution of an electron in a uniform magnetic field grows quickly with its energy, so a cyclotron cannot accelerate electrons. A microtron gets round this: the electron circulates in a uniform magnetic field and crosses a single accelerating gap once per revolution, and its orbits, of growing radius, all touch one another at that gap. The machine is tuned so that the period of revolution grows, from one revolution to the next, by $\Delta T$ equal to a whole multiple of the period $T_0$ of the accelerating field. Taking $\Delta T = T_0$ , a magnetic induction $B = 107$ mT and an accelerating field of frequency $\nu = 3000$ MHz, find how many times the electron must cross the gap to acquire the energy $W = 4.6$ MeV.

A microtron seen along its magnetic field. Every orbit touches the others at the accelerating gap on the right, fed from the oscillator ac; the outermost path leaves through the exit channel at the upper left.

Given

$W = 4.6$ MeV
$\Delta T = T_0$
$B = 107$ mT
$\nu = 3000$ MHz

Find

the number

n

of gap crossings

Solution

Step 1The period of revolution at any energy

For a circle of radius $r$ in the field $B$ the momentum is $p = eBr$ , and the speed is $v = pc^2/(mc^2 + W)$ , where $W$ is the electron’s kinetic energy at that stage and $mc^2 + W$ its total energy. One revolution takes the circumference over the speed:

T = \frac{2\pi r}{v} = \frac{2\pi\,(p/eB)}{pc^2/(mc^2 + W)}

T = \frac{2\pi\,(mc^2 + W)}{eBc^2}

(1)

The period is proportional to the total energy, which is why it grows as the electron is accelerated.

Step 2The energy gained at one crossing

Between two successive revolutions the electron crosses the gap once and its energy grows by $\Delta W$ . By (1) the period then grows by $\Delta T$ , and the tuning sets this equal to $T_0 = 1/\nu$ :

\Delta T = \frac{2\pi\,\Delta W}{eBc^2} = T_0 = \frac{1}{\nu}

Solving for the gain per crossing:

\Delta W = \frac{eBc^2}{2\pi\nu} \tag{2}

Every crossing must give this same amount, or the electron would fall out of step with the field.

Step 3The number of crossings

Starting from rest, the electron collects $\Delta W$ at each of $n$ crossings, so $n\,\Delta W$ equals the final kinetic energy $W = 4.6$ MeV. With (2):

n = \frac{W}{\Delta W} = \frac{W}{eBc^2/(2\pi\nu)}

Numerically, with $W = 4.6$ MeV $= 7.36\times10^{-13}$ J and $c = 3.00\times10^{8}$ m/s,

n = \frac{2\pi\,(3.0\times10^{9})(7.36\times10^{-13})}{(1.60\times10^{-19})(0.107)(3.00\times10^{8})^2} = 9.0

n = \frac{2\pi\nu W}{eBc^2} = 9

(3)

Check

Units: $eBc^2$ is $\text{C}\cdot\text{T}\cdot\text{m}^2/\text{s}^2$ , and with $\text{T} = \text{kg}/(\text{C}\cdot\text{s})$ it is $\text{kg}\cdot\text{m}^2/\text{s}^3$ , an energy per second; multiplied by $1/\nu$ it is an energy, so $\nu W/(eBc^2)$ is a pure number. The gain per crossing from (2) is $1.54\times10^{-3}/(2\pi\cdot3.0\times10^{9})$ J $= 8.2\times10^{-14}$ J $= 0.511$ MeV, and $4.6/0.511 = 9.0$ , a whole number, as a count of crossings must be.

Remark

The gain per crossing comes out equal to the electron’s rest energy $mc^2$ for these numbers, and that is the design: with $\Delta W = mc^2$ , the first orbit (total energy $2mc^2$ ) has period $2\cdot 2\pi m/(eB)$ , and $2\pi m/(eB) = 0.334$ ns matches the period $T_0 = 0.333$ ns of the 3000 MHz field to 0.2 per cent. Every orbit then lasts a whole number of field periods, so the electron always meets the gap in the accelerating phase, the condition a cyclotron cannot meet once the electron becomes relativistic.