Answer:
The radius of the gold nucleus is 7.1x10⁻¹⁵m
Explanation:
The nearest distance is:
[tex]r=\frac{kze^{2} }{mpV^{2} }[/tex] (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:
[tex]E_{k} =\frac{1}{2} m V^{2}[/tex]
Where
m = 4 mp = mass of proton
[tex]5.12x10^{-12} =\frac{1}{2} *4*m_{p}* V^{2} \\m_{p}* V^{2} = 2.56x10^{-12}[/tex]
Replacing in equation 1
[tex]r=\frac{9x10^{9}*79*(1.6x10^{-19})^{2} }{2.56x10^{-12} } =7.1x10^{-15} m[/tex]
