Wavelength of Hard $\gamma$-Rays

A. H. Compton. The Wave-Length of hard Gamma Rays. Phil. Mag. 41 p. 770; 1921.

The theory of a simple optical grating for monochromatic light shows that, whereas the distance between spectra of different orders is determined by the so-called grating constant $a+b$, where $a$ is the width of the slit and $b$ the distance between adjacent slits, in the diffraction phenomenon consisting of bands the distance between the latter depends only on the width of the slit. In the first case we have the formula

\[ \sin \varphi=\frac{n\lambda}{a+b}, \]

where $\lambda$ is the wavelength, $n$ the order of the spectrum, and $\varphi$ the angle between the incident and the deflected ray. In the second case the positions of the maxima for the bands are successively determined by the following equalities:

\[ \sin \varphi=0;\quad \frac{1.430\lambda}{a};\quad \frac{2.459\lambda}{a}\quad \text{etc.} \]

It is evident from this that the angular distance between the bands is considerably greater than for lines.

From measurements of the wavelengths of $\gamma$-rays from Ra C by Rutherford and Andrade in 1914, the smallest value obtained for $\lambda$ was $0.07 \cdot 10^{-8}$ cm. Meanwhile, the $\beta$-rays of the same element have velocities corresponding to from 5 to 20.10, which gives for $\lambda$ for the hardest $\gamma$-rays a value of about $0.007 \cdot 10^{-8}$ cm. Moreover, from the formula relating the absorption coefficient of X-rays to the wavelength, knowing the absorption coefficient for $\gamma$-rays, it may be calculated that the lower limit for $\lambda$ will be $0.04 \cdot 10^{-8}$ cm.

This casts doubt on the results obtained by Rutherford and Andrade, and therefore Compton proposes another method for determining the wavelength of $\gamma$-rays. Rutherford and Andrade used reflection from the internal faces of an NaCl crystal, i.e. a crystal lattice; Compton uses an “atomic grating.” The point is that the orientation of the electrons in the atom creates a peculiar diffraction pattern, depending on the wavelength and on the dimensions of the atom.

The diffraction pattern from a crystal lattice is determined by the distance between the planes that produce reflection, and in the case of hard $\gamma$-rays it is all contained within a very small angle (about $1^\circ$). The phenomenon of diffraction from an atom, however, is spread over a wide angle ($>10^\circ$), since here $a$ (the radius of the atom) is very small in comparison with the grating constant. These two phenomena will be quite analogous to the line and band spectra from a simple grating in the optics of visible rays.

Since the radius of the atom is unknown to us, in determining the wavelength one must resort to an indirect method, namely to comparing the ratio of the intensities of $\gamma$-rays scattered by atoms of any two elements (Compton takes copper and lead) with the ratio of the intensities of scattered X-rays for the very same elements. Debye’s theory for the scattering of X-rays gives the following expression for the ratio of the intensity of scattered rays $J_{\theta}$ to the intensity of the incident rays $J$:

\[ \frac{J_{\theta}}{NJ}=\psi \frac{\sin \theta/2}{\lambda}, \]

where $N$ is the atomic number and $\theta$ the angle between the incident and the scattered ray.

From the data of Barkla and Dunlop, obtained from the scattering of X-rays at an angle of $90^\circ$, one can determine the ratio of the intensities for lead and copper for

\[ \frac{\sin 45^\circ}{\lambda} \]

where $\lambda$ varied from $0.3 \cdot 10^{-8}$ to $1.10^{-8}$ cm. It is evident that, if the same

the ratio of intensities is obtained for an unknown wavelength and at a definite angle \(\theta'\), then we may write

\[ \frac{\sin \dfrac{\theta'}{2}}{\lambda_{\infty}} = \frac{\sin \dfrac{\theta}{2}}{\lambda}, \]

whence

\[ \lambda_{\infty}=\lambda\, \frac{\sin \dfrac{\theta'}{2}}{\sin \dfrac{\theta}{2}}, \]

where

\[ \frac{\theta}{2}=45^\circ. \]

Compton, investigating the scattering of \(\gamma\)-rays that had passed through an 8 mm thickness of lead, determined that for an angle of \(10^\circ\) there is obtained a ratio close to that from the data of Barkla and Dunlop for

\[ \frac{\sin 45^\circ}{0.25\cdot 10^8}. \]

Consequently, we have

\[ \frac{\sin 45^\circ}{0.25\cdot 10^8} = \frac{\sin 5^\circ}{\lambda_{\infty}}, \]

whence

\[ \lambda_{\infty}=0.03\cdot 10^{-8}\ \mathrm{cm}. \]

Compton’s method cannot give exact results, above all because the measurement of the intensities of the scattered rays (separating the scattered rays from the rays of fluorescence) presents great difficulties. Moreover, Compton’s data do not overlap with those of Barkla and Dunlop, and as a result it is necessary to extrapolate them to those values of the intensity ratios which were obtained by Compton.

N. Selyakov.