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A nuclear raction is given us P+^(15)N rarr _(Z)^(A)X +n (a). Find, A,Z and identitythe 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 momentumof 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."]"):}. |
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Answer» (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^(@)` . |
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