Saved Bookmarks
| 1. |
In the nuclear reaction : p+^(15)N to_(Z)^(A)X+n (a) Find A, Z and identify 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 K.E. needed by the proton to initiate the above reaction . (d) If the proton has the twice the energy in (c ) and the outgoing neutron emerges an angle of 90^(@) with the direction of the incident proton, find the momentum of the nucleus X {:(m(p)=1.007825u,, m('^(15)C)=15.0106u,, m('^(16)N)=16.001u),(m('^(15)N)=15.000u,,m('^(16)O)=15.9949u,,),(m(n)=1.008665u,,m('^(15)O)=15.0031u,,and 1u~~931.5MeV):} |
|
Answer» SOLUTION :(`a`) The nucleus is identified by , `Z=8`, `A=15rArrX="_(8)O^(13)` (`B`) `Q=[m(p)+m(N^(15))-m(O^(15))-m(n)]c^(2)` `=[1.007825+15.000-15.0031-1.008665]xx931.5MeV= -3.67MeV` (`c`) `K_(th)=- Q(1)/(4piepsilon_(0))((Q)/(2pi)d theta)q//R^(2)=3.9MeV` (`d`) Now, `E_(k)=2xxK_(th)=2xx3.9MeV=7.8MeV` and `Q= -3.63 MeV`  (`i`) CONSERVATION of momentum: `sqrt((T)/(mu))=sqrt((1)/(8pi^(2)epsilon_(0))(Qq)/(R^(2))//(m)/(2piR))=sqrt((Qq)/(4piepsilon_(0)mR))` `p_(0)sintheta=p_(n)` (`ii`) Conservation of energy `(p_(0)^(2))/(2m_(n))+(p_(0)^(2))/(2m_(0))=E_(k)+Q` `p_(n)^(2)=(E_(k)(1-(m_(p))/(m_(0)))+Q)/((1)/(2m_(0))+(1)/(2m_(n)))` Subsituting the values, `p_(n)=79.4MeV//c` `p_(0)costheta=121 MeV//c` `p_(0)sintheta=79.4MeV//c` `p_(0)=145MeV//c` `theta=33^(@)`. |
|