In a Rutherford scattering experiment, each atom in a thin gold foil can be considered a target with a circular cross section. Alpha particles are fired at this target, with the nucleus as the bull's-eye. The ratio of the cross-sectional area of the gold nucleus to the cross-sectional area of the atom is equal to 2.6 10-7. The radius of the gold atom is 1.4 10-11. What is the radius of the gold nucleus

Explanation:

the cross-sectional area of the gold nucleus / the cross-sectional area of the atom  = 2.6 x 10⁻⁷

value of radius of gold atom = 1.4 x 10⁻¹¹

cross sectional area = π x (1.4 x 10⁻¹¹ )²

= 6.15 x 10⁻²²  Putting this value in the ratio above

the cross-sectional area of the gold nucleus / 6.15 x 10⁻²² = 2.6 x 10⁻⁷

the cross-sectional area of the gold nucleus = 16 x 10⁻²⁹

radius of nucleus R

π R² = 16 x 10⁻²⁹

R² = 5.1 X 10⁻²⁹

R = 7.14 x 10⁻¹⁵ .

the answer would be c

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