Saved Bookmarks
| 1. |
A positron with kinetic energy T= 750 keV strikes a stationary free electron. As a result of annihilation, two gamma quanta with equal energies appear. Find the angle of divergence between them. |
|
Answer» Solution :By momentum conservation `sqrt(E^(2)-m_(e )^(2)C^(4))=2(E+m_(e )c^(2))/(2)"cos" (THETA)/(2)` or `"cos "(theta)/(2)= sqrt((E-m_(e )c^(2))/(E+m_(e )c^(2))) =sqrt((T)/(T+2M_(e )c^(2)))` Substitution GIVES `theta= 98.8^(@)` 
|
|