X-rays of wavelength 22 pm are scattered from a carbon target at an angle of ` 85^(@)` to the incident beam The Compton shift for X-rays is ` (cos 85^(@) = 0.088)`

Explanation

The nuclear reaction in question can be written as-`P_(1)^(1) +HE_(1)^(3)+n_(0)^(1)+Q`,hwn Qis the energy realsed during the reaction given by `Q=[(m_p+m_(H))-(m_(He)+m_(n))]`amu =(1.0072765+3.06056)-(3.016030-1.008665)=-0.001369amu=-1.2745 MeV,From conservation of energy ,we get `K_(p)=K_(n)+K_(He)+Q`,`implies K_(n)=K_(H)-K(He)-Q=3 -K_(He)+1.2745=4.2745-K_(He)`.....(1),From conservation of linear momentum along and perpendicular to the direction of proton beam ,we get `p_(H)=p_(n)cos 30^(@)+p_(He)costheta implies p_(H) =(sqrt 3)/(2)p_(n)+p_(He)cos theta......(2) and `0=P_(n)sin30^(@)-p_(He)sin theta` `implies p_(He)sintheta=(p_(n))/(2)`.......(3)Using the formula for kinetic energy `k=(p^(2))/(2m)`in the above equations (1)-(3), and puttting the values of masses from the given data, we get , `implies K_(n)=1.444MeV`.