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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?

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No, the gamma ray doesn't strike the nucleus. You need something around to make it possible to conserve momentum and energy and to obey the relationship between them for every particle. In sum, there is no way to make two massive objects, like the electron and positron, move that will make the pair of them act like they have zero mass the way a photon does.

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.

In the case of the nucleus, the nucleus supplies a virtual photon, one that doesn't obey (1) and so cannot exist for very long. So, the gamma ray does not strike the nucleus, directly, but the nucleus does recoil under the force supplied by a virtual photon, and the virtual photon supplies the needed energy and momentum to make a virtual electron or positron real.

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