`C^(14)` disintegrates by `beta`-emission with a reaction eneregy (Qvalue) of `0.155 MeV`
Channel: Doubtnut
Date: 26/07/2020

`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))` .