To deduce the expected intensity of scattered gamma rays from the aluminum target, you need to know the total number of electrons in the target. The target is a cylinder of height h and diameter d. Aluminum has atomic number 13 (13 electrons per atom), atomic mass of 26.9815 g/mole, and mass density of 2.7 g/m3 CTn.

1. If the cylindrical target has a height of h=1.91 cm and a diameter of d=1.91 cm, the volume of the target is V = cm^{3}.

2. Suppose the actual volume of the target is V=5.387 cm^{3}. The total number of electrons in this aluminum target is: N = x10^{24} electrons.