Question

The predictions of Rutherford's scattering formula failed to correspond with experimental data when the energy of the incoming alpha particles exceeded 32MeV. This can be explained by the fact that the predictions of the formula apply when the only force involved is the electromagnetic force and will break down if the incoming particles make contact with the nucleus. Use the fact that Rutherford's prediction ceases to be valid for alpha particles with an energy greater than 32MeV to estimate the radius r of the gold nucleus. I am doing something really silly. I keep getting 7.43*10^-31 which is off by a factor of 16. I need a detailed explaination so I can find out where I went wrong. I equated the KE = 5.12e-12 = k(79e^2/r)

Answer 1

Answer:

The radius of the gold nucleus is 7.1x10⁻¹⁵m

Explanation:

The nearest distance is:

$r=\frac{kze^{2} }{mpV^{2} }$ (eq. 1)

Where

z = atomic number of gold = 79

e = electron charge = 1.6x10⁻¹⁹C

k = electrostatic constant = 9x10⁹Nm²C²

energy of the particle = 32 MeV = 5.12x10⁻¹²J

At the potential energy is zero, all the energy will be kinetic energy:

$E_{k} =\frac{1}{2} m V^{2}$

Where

m = 4 mp = mass of proton

$5.12x10^{-12} =\frac{1}{2} *4*m_{p}* V^{2} \\m_{p}* V^{2} = 2.56x10^{-12}$

Replacing in equation 1

$r=\frac{9x10^{9}*79*(1.6x10^{-19})^{2} }{2.56x10^{-12} } =7.1x10^{-15} m$