# In pair production does gamma rays actually strike a nucleus?

I think the quetion is vagueless.I simply want to ask it is often printed that there must be a heavy nucleus present for pair production.It is answered that its needed to make electron and positron existable after prodution.Major question in that gamma rays actually strike nucleus for pair production?If not then how is pair formed?

See, for a photon the momentum and energy obey the relationship $$E_\gamma = p_\gamma c. \tag1$$ For an electron and positron, they obey $$E_{\pm}^2 = (m_e c^2)^2 + (p_{\pm}c)^2. \tag2$$ Now, the conservation laws require that the total momentum and energy are unchanged, so \begin{align} E_\gamma &= E_+ + E_-,\ \mathrm{and} \tag3\\ \vec{p}_\gamma &= \vec{p}_+ + \vec{p}_- . \tag4 \end{align} If you square (3) and $c$ times (4) and subtract them you get $$E_\gamma^2-p_\gamma^2c^2 = E_+^2 + 2 E_+E_-+E_-^2 - c^2p_+^2 - 2 c^2\vec{p}_+\cdot\vec{p}_- - c^2p_-^2. \tag5$$ Applying (1) and (2) (the mass shell relations) to (5) gives $$0 = (m_e c^2)^2+ \sqrt{\left[(m_ec^2)^2+(p_+c)^2\right]\left[(m_ec^2)^2+(p_-c)^2\right]} - c^2p_+p_-\cos\theta. \tag6$$ What you'll find is that no matter what real values of $p_+$, $p_-$, and $\theta$ you put in to (6) you cannot get the right hand side to equal the left.
In order to end up with something that has non-zero $(E_++E_-)^2-\left(\vec{p}_+c +\vec{p}_-c\right)^2$ you need $(E_1+E_2)^2-\left(\vec{p}_1c +\vec{p}_2c\right)^2$ to be non-zero to begin with. The main way to accomplish that is to have another photon moving in roughly the opposite direction before the collision.