# Elastic Proton Scattering on Nucleus (Nuclear Physics)

A proton  recoils at an angle $alpha$, which equals $60$ degrees, when is scattered elastically at an angle of $56$ degrees by a nucleus. What is its atomic mass and what fraction of kinetic energy is transferred to the nucleus.

Take $m_1= 1$ mass of proton, $m_2$ mass of nucleus, $v_1$ incoming speed of proton, $v_1’$ scattered speed of proton, $v_2’$ scattered speed of nucleus initially at rest ($v_2=0$). One can write for momentum on x,y axes and for the energy (for elastic collision energy is conserved):

$m_1 v_1′ sin(56)=m_2 v_2′ sin⁡(60)$

$m_1*v_1′ cos⁡(56)+m_2*v_2′ cos⁡(60)=m_1 v_1$

$0.5*m_1*v_1^2=0.5*(m_1*v_1’^2+m2*v_2’^2)$

Apply a small trick: take $m_1=1$ (proton) and $v_1 =1$ (normalize at 1 the incoming speed, the results will be the same):

$v_1’*cos⁡(56)+m_2*v_2’*cos⁡(60)=1$

$v_1’^2+v_2’^2=1$

$v1’*sin⁡(56)=m_2*v_2’*sin⁡(60)$

Divide (3) by (1) rearranged to get:

$tan⁡(60)=(v_1’*sin⁡(56))/(1-v_1’*cos⁡(56))$,that is

$v_1’=(tan⁡(60))/(sin⁡(56)+cos⁡(56)tan⁡(60))=0.9635$

And also

$tan⁡(56)=(m_2*v_2’*sin⁡(60))/(1-m_2*v_2’*cos⁡(60))$

or $m_2*v_2’=(tan⁡(56))/(cos⁡(60) tan⁡(56)+sin⁡(60))=0.9224$

From (2) one has

$v_2’=sqrt{(1-v_1’^2)}=0.2677$

Which gives

$m_2=0.9224/0.2677=3.446$

And

$(m_2*v_2’^2)/(m_1*v_1^2 )=3.446*0.2677^2=0.2469$