Answer to Question #105440 in Molecular Physics | Thermodynamics for Sophia

Question

In a router for scattering experiment and alpha particle is positive charge 2 is directed towards a gold necklace with charms 79-80 alpha particle has a kinetic energy of 5.0 even went very far from the nucleus as even the Golden close to be fixed and space determine the distance of the closest approach note 8.99 * 10 is 4 and 9 Newton per hour metre square column column

Answer 1

From the conservation of energy:

$$K=\frac{kq_1q_2}{r}$$

$$5(1.6\cdot 10^{-13})=\frac{(8.99\cdot 10^{9})(2)(79)(1.6\cdot 10^{-19})^2}{r}$$

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$$r=4.5\cdot 10^{-14}m$$