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Explain Rutherford's explanation for scattered alpha-particles. |
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Answer» Solution :Rutherford suggested that since large number of a-particles are scattered at very small angles, atoms must be LARGELY hollow. Since the gold foil is very thin, it can be assumed that `alpha` particles will suffer not more than one scattering during their PASSAGE through it. `alpha`-particles are nuclei of helium atoms and CARRY two + 2e charge and have the mass of the helium atom. For gold Z = 79, the nucleus of gold is about 50 times heavier than an a-particle it is assume that stationary THROUGHOUT the scattering process. Under these assumptions, the trajectory of an a-particle can be computed using Newton.s second law of motion and the Coulomb.s law for force of repulsion between the O-particle and the positively charged nucleus. The MAGNITUDE of Coulomb.s repulsive force `I^(2)=(1)/(4pi epsi_(0)) *((Ze)(2e))/(r^(2))` wherer the distance between the c-particle and the nucleus and `epsi_(0)`is the permittivity of vacuum. The force is directed along the line joining the cparticle and the nucleus and varies continuously with the displacement of `alpha`-particle |
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