contestada

What fraction of the α particles in Rutherford's gold foil experiment are scattered at large angles? Assume the gold foil is two layers thick, as shown in the figure, and that the approximate diameters of a gold atom and its nucleus are 1.7 Å and 1.0×10−4Å , respectively. Hint: Calculate the cross-sectional area occupied by the nucleus as a fraction of that occupied by the atom. Assume that the gold nuclei in each layer are offset from each other.

Respuesta :

pstnonsonjoku

The fraction of the  cross-sectional area occupied by the nucleus as a fraction of that occupied by the atom is 3.48 * 10^-9 .

What is the cross sectional area of the nucleus?

Now we know that the atom is composed of the nucleus that houses the protons and the neutrons while the electrons are found on the shells.

We have the information in the question that the diameter of the atom is given as  1.7 Å or 1.7 * 10^-10 m and the diameter of the nucleus is 1.0×10−4Å or 1.0×10^−14 m.

Area of the atom = 3.142 * ( 1.7 * 10^-10/2)^2 = 2.27 * 10^-20 m^2

Area of the nucleus =  3.142 * (1.0×10^−14/2)^2 = 7.9 * 10^-29 m^2

The fraction of the  cross-sectional area occupied by the nucleus as a fraction of that occupied by the atom is ; 7.9 * 10^-29 m^2 /2.27 * 10^-20 m^2

= 3.48 * 10^-9

Learn more about Rutherford:https://brainly.com/question/4593249

#SPJ1

Otras preguntas