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.

Answer

(a) `eta=4mM//(m+M)^(2)=0.89`
(b)`eta=2m//(m+M)=2//3`
Here `m` and `M` are the masses of a neutron and deutron.

Explanation

According conservation principal of momentum,
`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)`