A nuclear raction is given us
` P+^(15)N rarr _(Z)^(A)X +n`
(a). Find, `A,Z` and identity the nucleus `X`. (b) Find the `Q` value of the reaction . ( c) If the proton were to collide with the `.^(15)N` at rest, find the minimum KE needed by the proton to initiate the above reaction. (d) If the proton has twice energy in (c) and the outgoing neutron emerges at an angle of `90^(@)` with the direction of the incident proton, find the momentum of the protons and neutrons. `{:("[Given,"m(p)=1.007825 u","m(.^(15)C)=15.0106u","),(m(.^(16)N)=16.0061 u","m(.^(15)N)=15.000u"),"),(m(.^(16)O)=15.9949 u","m(u)=1.0086665u ","),(m(.^(15)O)=15.0031u"," and 1 u ~~ 931.5MeV."]"):}`.

Answer

(a) `Z=8,A =15 rArr X =._(8)O^(15)`, (b) `-3.67 MeV`, (c ) `3.9MeV`, (d) `P_(n)=79.4 MeV//c`

Explanation

(a) The nuleus is identify by:`Z=8,A=15 rArr X =_(8)O^(15)`
(b) `Q=[m(P)+(m)(N^(15))-m(O^(15))-m(n)]c^(2)=-3.67 MeV`
(c ) `K_(th)= -Q(1+(m_P)/(m_(M)))=3.9 MeV`
(d) Now,`E_(k) =2 xx K_(th) =2 xx 3.9 MeV=7.8 MeV` and `Q = -3.63 MeV`
(a) Conservation of momentum:
`p_(0)cos theta =sqrt(2m_(p)E_(k))`
`P_(0) sin theta =p_(n)` (b) Conservation of energy:
`(p_(n)^(2))/(2m_(n))+(p_(0)^(2))/(2m_(0))=E_(k)+W`
`p_(n)^(2)=(E_(k)(1-(m_(p))/(m_(0)))+Q)/((1)/(2m_(0))+(1)/(2m_(n)))`
`:. P_(n) =79.4 MeV//c,P_(0)=145 MeV//C` and `theta =33^(@)` .