Theory of Dopler Broadening

Photon emmision along direction of fission fragment velocit

= amount of energy the gamma ray was shifted by

where $E_{photon}$ is the Doppler-shifted $\gamma$ -ray energy which was measured

$cos\theta_{photon1}$ =- $cos\theta_{1}$ for an excited fragment in det-1
$cos\theta_{photon2}$ =- $cos\theta_{2}$ for an excited fragment in det-2

$\theta_i$ is the polar emission angle of excited fragment and $\beta_i$ the fragment velocity

$\beta_i$ =0.04634 $\sqrt{\frac{E_i[MeV]}{A_i[u]}}$

This picture is taken from Smith, A. B.; Fields, P. R.; Friedman A. M., Physical Review, vol. 104, Issue 3, pp. 699 - 702