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Relativistic momentum of a particle with rest mass $m$ and speed $v$ is
$p=mv$ $p=\gamma mv$ $p=mc$ $p=\gamma mc$ Photon momentum can be written as
$p=mc$ $p=E/c=h/\lambda$ $p=hf/c^2$ $p=\lambda/h$ Compton wavelength shift for scattering angle $\theta$ is
$\Delta\lambda=\dfrac{h}{m_ec}(1-\cos\theta)$ $\Delta\lambda=\dfrac{h}{m_ec}(1+\cos\theta)$ $\Delta\lambda=\dfrac{hc}{m_e}(1-\cos\theta)$ $\Delta\lambda=\dfrac{h}{m_e}(1-\cos\theta)$ Minimum photon energy for electron-positron pair production (near a nucleus) is approximately
$0.511\,\text{MeV}$ $1.022\,\text{MeV}$ $2.044\,\text{MeV}$ $938\,\text{MeV}$ An electron and positron annihilate from rest into two photons. The photons are
one photon of $1.022\,\text{MeV}$ two photons of $511\,\text{keV}$ in opposite directions two photons of different energies in same direction a single gamma ray and a neutrino Relativistic de Broglie wavelength of a massive particle is
$\lambda=\dfrac{h}{\gamma mv}$ $\lambda=\dfrac{h}{mv}$ always $\lambda=\dfrac{\gamma h}{mv}$ $\lambda=\dfrac{h\gamma v}{m}$ An electron and a photon have the same momentum magnitude. Which statement about their de Broglie wavelengths is correct?
They have the same wavelength because $\lambda=h/p$ for both. The photon has the longer wavelength because it is massless. The electron has the longer wavelength because it has rest mass. Their wavelengths cannot be compared without knowing both energies. Relativistic kinetic energy is
$K=\gamma m_0c^2$ $K=(\gamma-1)m_0c^2$ $K=\tfrac12 m_0v^2$ exactly $K=pc$ for all particles A muon has proper lifetime $\tau_0$. In the lab where it moves with Lorentz factor $\gamma$, mean lifetime is
$\tau_0/\gamma$ $\gamma\tau_0$ $\tau_0$ $\gamma^2\tau_0$ For a massless particle, the correct dispersion relation is
$E=pc$ $E=m_0c^2$ $E=\tfrac12 mv^2$ $p=m_0v$ Compared with classical prediction, Compton scattering confirms that X-rays carry
charge but no momentum momentum in discrete photon interactions only wave phase with no particle behavior rest mass equal to electron mass A free electron in vacuum cannot absorb a single photon by itself because
charge conservation fails energy conservation alone cannot be satisfied for any final momentum simultaneous energy and momentum conservation cannot both be satisfied photons have no momentum