Partial and Complete Transfer of Energy in Bremsstrahlung Must Include Spin and Vibrational Kinetic Energies

When complete braking is achieved, the spin and vibrational kinetic energies as well as linear kinetic energy is transferred to the resulting photon: $h\nu = 1/2mv^2 + 1/2I\omega_r^2 + (n +1/2)\hbar\omega_v$ If partial transfer of kinetic energy is achieved by decelerating a charged particle, then the resulting photon is $[1/2mv^2_2 +1/2I\omega_{r2}^2 + (n+1/2)\hbar\omega_{v2}]- [1/2mv_{1}^2 +1/2I\omega_{r1}^2 + (n+1)\hbar\omega_{v1}]$. $1/2I\omega_r^2$ is the spin kinetic energy and $(n +1/2)\hbar\omega_v$ is the vibrational kinetic energy. By using the spin and vibrational factors some reconciliation of experimental and theoretical values can be achieved.