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Find the gratest possible angle through which a deutron is scattered as a result of elastic collision with an initially stationary proton. |
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Answer» Solution :From CONSERVATION of momentum `sqrt(2MT)hat(i)=vec(p)_(d)+vec(P)_(p)` or `P_(p)^(2)= 2MT+p_(d)^(2)-2sqrt(2MT)p_(d) COS THETA` From energy conservation `T=(p_(d)^(2))/(2M)+(p_(p)^(2))/(2m)`  `(M=` MASS of denteron, `m=` mass of proton) So `P_(p)^(2)= 2mT-(m)/(M)p_(d)^(2)` Hence `p_(d)^(2)(1+(m)/(M))-2 sqrt(2MT) p_(d) cos theta+2(M-m)T=0` For real roots `4(2MT) cos^(2) theta-4xx2(M-m)T(1+(m)/(M)) ge0` `cos^(2) theta ge(1-(m^(2))/(M^(2)))` Hence `sin^(2)theta le(m^(2))/(M^(2))` i.e., `theta le"sin"^(-1)(m)/(M)` For deuteron-proton scalterting `theta_(max)= 30^(@)`. |
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