Question

There are Z protons in the nucleus of an atom, where Z is the atomic number of the element. An α particle carries a charge of +2e. In a scattering experiment, an α particle, heading directly toward a nucleus in a metal foil, will come to a halt when all of the particle's kinetic energy is converted to electric potential energy. In such a situation, how close will the center of an α particle with a kinetic energy of 6.4 x 10-13 J come to the center of a gold nucleus (Z = 79)?

Answer 1

Answer:

$r = 2.84 \times 10^{-14} m$

Explanation:

As per energy conservation we know that the electrostatic potential energy of the charge system is equal to the initial kinetic energy of the alpha particle

So here we can write it as

$\frac{1}{2}mv^2 = \frac{k(2e)(ze)}{r}$

now we know that

$m = 1.67 \times 10^{-27} kg$

$e = 1.6 \times 10^{-19} C$

z = 79

here kinetic energy of the incident alpha particle is given as

$KE = 6.4 \times 10^{-13} J$

now we have

$6.4 \times 10^{-13} = \frac{(9\times 10^9)(1.6 \times 10^{-19})^2(79)}{r}$

now we have

$r = 2.84 \times 10^{-14} m$