Full Text
Polarization of Bremsstrahlung
Usually, in the study of various $\gamma$-reactions on electron accelerators, it is assumed that the beam of $\gamma$-rays arising from the braking of electrons in the target is unpolarized.
Calculations$^{1,2}$ carried out for electron energies
\[ E_0 \gg 137 Z^{-1/3} mc^2, \]
where $Z$ is the atomic number of the target, show that a predominant polarization of bremsstrahlung should be observed in the direction perpendicular to the plane formed by the directions of the primary electron and the emitted photon (the emission plane).
The degree of polarization is a function of the product of the electron energy $E_0$ and the angle $\theta$ between the directions of the incident electron and the photon. The optimum angle is
\[ \theta \sim \frac{mc^2}{E_0}. \]
In addition, the degree of polarization $(pp)$ depends on the value
\[ \frac{h\omega}{E_0}; \]
moreover, at the boundaries of the spectrum (at $h\omega = 0$ and $h\omega = E$), $pp = 0$.
However, in the same work it is shown that the polarization effect should be strongly weakened owing to multiple scattering of electrons in the target. It has been calculated that, for target thicknesses of 0.0096, 0.0072, and 0.0015 mm, the percent polarization is, respectively (for $E_0 = 1$), 3, 10, and 42.
In the paper under review$^{3}$ an attempt was made to detect the polarization of the $\gamma$-beam from a betatron with maximum energy $E = 20.0$ MeV. Of a number of methods for recording the polarization process (observation of the formation of electron–positron pairs, formation of photoelectrons, etc.), the author used a method based on the use of the photodisintegration reaction of the deuteron*). The protons formed as a result of the reaction fly out in the direction of the electric vector. A thin target (10 microns thick) of heavy paraffin $(\mathrm{C_nD_{2n+2}})$, deposited by evaporation on a piece of cellophane, was placed at the center of the collimated lead beam (Fig. 1). The half-width of the beam at the target was about $3^\circ$. Immediately behind the target an Ilford C-2 photographic plate with an emulsion of thick-
Fig. 1. Schematic arrangement of the apparatus: $OAB$ is the emission plane, $OA$ is the direction of the accelerated electrons, $OB$ is the direction of the photon.
*) With a similar method, polarization of $\gamma$-rays formed in the reaction $\mathrm{H^2}(p\gamma)\mathrm{He^3}$ was found.$^{4}$
of 200 microns. To avoid blackening of the emulsion, the exposure during irradiation was only 1 minute. Four regions of the plate were examined, corresponding to different angles \(AOB\). For each track, the angle was found between the plane of emission and the projection of the track of a proton with energy from 6 to 15 MeV onto a plane perpendicular to the plane of emission. The distribution, obtained for 500 tracks, of the number of photoprotons as a function of the angle is presented in Fig. 2. The maximum observed at \(90^\circ\)
Fig. 2. Distribution of photoprotons.
qualitatively confirms the conclusions of the theories, although the large thickness of the betatron target (\(0.125\) mm) should have led to a practically unobservable magnitude of the polarization because of the large amount of multiple scattering of the electrons in the target.
The author also finds no explanation for the existence of the second maximum at \(20^\circ\).
B. R.
References
- M. May, Phys. Rev., 84, 265 (1951).
- M. May and Wick, Phys. Rev., 81, 628 (1948).
- K. Phylips, Phyl. Mag., 44, 141 (1953).
- Phyl. Mag., 43, 659 (1952).