The energy required to remove an electron from an atom is known as the ionization energy. The ionization energy for gold is 890 kJ/mol, or 1.48x10^-18 J per atom.
Calculation:
- The average kinetic energy of an alpha particle is given as 2x10^10 J.
- We can convert this to joules per atom by dividing by Avogadro's number, which is 6.02x10^23 atoms per mole. This gives us:
2x10^10 J / 6.02x10^23 atoms = 3.32x10^-14 J per atom
- We can then divide this energy by the ionization energy per atom to determine the maximum number of electrons that could be stripped from a gold atom by a single alpha particle:
3.32x10^-14 J per atom / 1.48x10^-18 J per atom = 2.24x10^4 electrons
- Finally, we can estimate the atomic radius of gold using the Bohr model by assuming that the outermost electrons in a gold atom are in the highest energy level. The radius of this energy level can be estimated using the Bohr radius formula:
r = n^2*h^2 / (4*pi^2*m*e^2)
where:
n = 3 (assuming the outermost electrons are in the third energy level)
h = Planck's constant = 6.626x10^-34 J*s
m = the mass of an electron = 9.11x10^-31 kg
e = the charge of an electron = 1.602x10^-19 C
Plugging in these values, we get:
r = 3^2*(6.626x10^-34 J*s)^2 / (4*pi^2*9.11x10^-31 kg*(1.602x10^-19 C)^2) = 1.36x10^-10 m
- Therefore, the maximum size of a gold atom that could be created by a single alpha particle collision is approximately twice the atomic radius, or 2.72x10^-10 m.
Hope this clarifies your question!