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4 January, 21:07

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

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  1. 4 January, 23:57
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    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.
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