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A neutron collides elastically with an initially stationary deuteron. Find the fraction of the kinetic energy lost by the neutron (a) in a head-on collision , (b) in scattering at right angles. |
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Answer» `VEC(p)_(1)+vec(p)_(2)=vec(p)_(3)+vec(p)_(4)` or `p_(1)=p_(3)+p_(4)`.....(4) According conservation principle of kinetic energy `T_(1)+T_(2)=T_(3)+T_(4) or T_(1)=T_(3)+T_(4)` or `(p_(1)^(2))/(2m)=(p_(3)^(2))/(2m)+(p_(4)^(2))/(2m)` or `(p_(1)^(2))/(2m)=((p_(1)-p_(4))^(2))/(2m)+(p_(4)^(2))/(2m)` (from eqn.(i)) or `(p_(1)^(2))/(2m)=(p_(1)^(2)+p_(4)^(2)-2p_(1p_(4)))/(2m)+(p_(4)^(2))/(2m) or (p_(1)^(2))/(2m)=(p_(1)^(2))/(2m)+(p_(4)^(2))/(2m)+(p_(4)^(2))/(2m)` or `p_(4)((1)/(m)+(1)/(M))=(2p_(1p4))/(2m) or p_(4)((1)/(m)+(1)/(M))=(2p_(1))/(m)` or `p_(4)=((2M)/(m+M))p_(1)` `:. p_(4)^(2)=((2M)/(m+M))^(2)p_(1)^(2)`.....(II) `:.` Loss in kinetic energy of neuton `=Delta T_(1)=` gain in `K.E`. of deuteron `DeltaT_(1)=T_(4)=(p_(4)^(2))/(2M)` FRACTION =`(DeltaT_(1))/(T_(1))=(((2M)/(m+M))^(2)(p_(1)^(2))/(2M))/((p_(1)^(2))/(2m))` `eta=((2M)/(m+M))((m)/(M))=(4mM)/((m+M)^(2))` Here `M~~2m` `:.` Fraction in `eta=(8)/(9)=0.89` (b) According to conservation priciple of momentum, `vec(p)_(1)+vec(p)_(2)=vec(p)_(3)+vec(p)_(4)` or `vec(p_(1))+0=vec(p)_(3)+vec(p)_(4)` or `vec(p)_(1)-vec(p)_(3)^(2)-2p_(1)p_(3)+vec(p)_(4)` Applying parallelogram law of vectors `p_(1)^(2)+p_(3)^(2)-2p_(1)p_(3)cos 90^(@)=p_(4)^(2)` According to conservation principle of kinetic energy `T_(1)+T_(2)=T_(3)+T_(4)` or `T_(1)+0=T_(3)+T_(4)` or `(p_(1)^(2))/(2m)=(p_(3)^(2))/(2m)+(p_(4)^(2))/(2m)` or `(p_(1)^(2))/(2m)=(p_(3)^(2))/(2m)+(p_(1)^(2)+p_(3)^(2))/(2M)` or `(p_(1)^(2))/(2m)=(p_(3)^(2))/(2m)+(p_(1)^(2)+p_(3)^(2))/(2M)` or `(p_(1)^(2))/(2m)=(p_(3)^(2))/(2m)+(p_(1)^(2)+p_(3)^(2))/(2M)` or `(p_(1)^(2))/(2m)((1)/(m)-(1)/(M))=(p_(3)^(2))/(1)((1)/(m)-(1)/(M))` or `p_(1)^(2)(p_(1)^(2))/(2m)((M-m)/(mN))= +p_(3)^(2)((m+M)/(mM))` or `p_(3)^(2)=((m-M)/(m+M))p_(1)^(2)` `:.` Loss in kinetic energy of NEUTRON is `DeltaT=T_(1)-T_(3)` `=(p_(1)^(2))/(2m)-(p_(3)^(2))/(2m)=(p_(1)^(2)-p_(3)^(2))/(2m)` `eta=(DeltaT)/(T_(1))=((p_(1)^(2)-p_(3)^(2))/(2m))/((p_(1)^(2))/(2m))` `1-(p_(3)^(2))/(p_(1)^(2))=1-((M-m)/(M+m))(p_(1)^(2))/(p_(1)^(2))` `h=1-((M=m)/(M+m))=(2m)/(M+m)=(2)/(3)` 
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