`C^(14)` disintegrates by `beta`-emission with a reaction eneregy (Qvalue) of `0.155 MeV` .A `beta`-particle with an energy of `0.025 MeV` is emitted in a direction at `135^(@)` to the direction of motion of the recoil nucleus. Determine the momneta of the three particles`(beta^(-)=barV,^(14)N)` involved in this disintegration in `MeV//c` units (where `c` is speed of light in vaccum)
`( M_(0) =0.511 MeV//c^(2))` .

Answer

`0.13,MeV//c,0.158 MeV//c,0.179 MeV//c`

Explanation

We know that the` Q` value of the reaction,
`E_(0)=E_(v)+E_(beta)`
`E_(v)=E_(0)-E_(beta)`
`=0.1555-0.025`
`=0.130 MeV`
Hence, the momentum of the neutrino
`P_(v)=E_(v//c)`
`=0.13 MeV//c`
Momentum of a `beta`-particle can be obtained from the relation
`P_(beta)=sqrt((2NM_(c)E_(beta)))=sqrt(2xx0.511xx0.025)MeV//c`
`=0.158 MeV//c`
Using conservation of momentum of the system, we have
`p_(beta)sin 45^(@)=p_(v) sin theta`
`sin theta =(p_(beta)//p_(v)) sin 45^(@)=0.859`
`p_(R) =P_(beta) cos 45^(@)+p_(v) cos theta`
`=0.158(1//sqrt2)+0.13 sqrt([1-(0.859)^(2)])`