![]() ![]() The second postulate, however, leads to a value for E m only one-fourth that given above, a value which is inconsistent with that derived from a consideration of radiation pressure and which leads to values for maximum ranges one-fiftieth of those observed by Wilson. The relative number of recoil electrons to photo-electrons increases with the frequency and is in agreement with observations by Wilson. The values for the maximum ranges in the case of x-rays (.34 to.48 A) are computed to be about one-third of those observed by Wilson for his fish tracts, but the difference may be due to the lack of homogeneity of the rays used. The maximum range increases rapidly with frequency. Wilson, the distribution of ranges is found to be such that two-thirds have tracks shorter than half the maximum range. On the first postulate the maximum recoil energy is shown to be E m = h ν 0 × 2 α ( 1 + 2 α ), where α = h mc λ 0 the recoil electrons are shown to be concentrated at angles near the direction of the primary beam and from the distribution of energy, using a relation given by C. Expressions for the distribution of recoil velocities, of energies and of ranges are developed for each of two postulates, assuming (1) the scattered radiation consists of directed quanta, and (2) the scattered radiation proceeds as spherical waves. Here I am stuck since this does not give me the required expression for the shift - just transforming frequencies to the corresponding wavelength does not give me the required result.Quantum theory of the recoil of electrons from scattered x-rays.-This is an extension of the quantum theory of scattering suggested by Compton, which assumes that each directed x-ray quantum is scattered by a single electron. Which after a little bit of algebra becomes $-$ Therefore from equations $(1)$ and $(2)$, we get $-$ Where $T$ is the energy given to the electron by the photon, which is essentially $h (\nu - \nu_0)$. Now we come to energy conservation: The total energy of the electron is given by $$p c \cos \phi = h \nu_0 + P c - h \nu \cos \theta$$ $$p c \sin \phi = h \nu \sin \theta$$ Where $\nu$ is the frequency of the electron after scattering, $p$ is the momentum of the electron after scattering, and $\phi$ is the electron scattering angle. Then momentum conservation gives, for the $x$- and $y$- axes respectively $-$ Suppose the photon and the electron are both moving initially along the $x$-axis. In proving this, I started in the same way as in the derivation for "stationary electron" - conservation of momentum and energy along each axis. $$\Delta \lambda = 2 \lambda_0 \frac$ is the initial energy of the electron. The book tells us that this shift is given by In a problem from Bransden and Joachain's Quantum Mechanics, it is asked to calculate the Compton wavelength shift, but the electron is now moving, with a momentum $P$, in the same direction as the approaching photon.
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