Find the minimum energy that a `gamma`-ray must have to give rise to an electron -positron pair.
`{:("Mass of electron",= 0.00055 am u,),("Mass of positron",= 0.00055 am u,):}`
Correct Answer - D
`(d)` We have `betaalpha+gammaalpha+alphabeta=0`
`Delta=(1)/(alpha^(3)beta^(3)gamma^(3))|{:(betagamma,gammaalpha,alphabeta),(gammaalpha,alphabeta,betagamma),(alphabeta,betagamma,gammaalpha):}|`
`=(1)/(alpha^(3)beta^(3)gamma^(3))|{:(betagamma+gammaalpha+alphabeta,gammaalpha,alphabeta),(gammaalpha+alphabeta+betagamma,alphabeta,betagamma),(alphabeta+betagamma+gammaalpha,betagamma,gammaalpha):}|` [using `C_(1)toC_(1)+C_(2)+C_(3)`]
`=(1)/(alpha^(3)beta^(3)gamma^(3))|{:(0,gammaalpha,alphabeta),(0,alphabeta,betagamma),(0,betagamma,gammaalpha):}|=0` [all zero property].