Question

a gamma beam 7 Mev hitted a Pb barrier and stop inside it , behind the barrier there was a Gamma Spectroscope.

As a result of this reaction the Spectroscope did't give the 7 Mev , instead gives a photonic Energy with value 511 Kev,

from these Answer the Question below .

1- what is the reaction that happened between the photon and the Pb barrier.

2- What is the The resulting particles from this reaction and what it's Energy.

3- if Photon with energy 511 Kev Hitted A Copper barrier , what the electron energy that in K orbit which will go as a resultant of absorbing 511 Kev if you know that the binding energy for electron is 8.98 Kev.

4- if the photon with 511 Kev scattered through an angle 45^o , what is the energy of scattered photon ?

Answer 1

A photon do interact with matter in a number of ways. These include photoelectric effect , compton effect, pair production etc.

When gamma ray of high energy of the order of MeV encounter the enter a high density medium like lead Pb it will attenuate appreciably due to high density.  

And also the gamma. Can then do pair production with energy aroun 1.02 MeVand can do compton scattering in relatively lower energy and photoelectric effect with photon of energy in eV. Also when it an electron produced encounter the Pb potential , Bremstralung radiations can also emitted.

Ultimately there is many possibilty of what comes out of the lead to the spectrometer and there is finite probabity of all of them

So the outcome can be a photon or and electron resulting due to compton scattering or pair production. And The energy of the particle that come out of lead is given in the question itself as 511 KeV.

I have given the reaction of pair production and compton effect below.

pair production cannot occur in m vaccum it happens in a presence of nucleus

answer to part A and B is included in above discussion

Answer to part. C and d