Ok, let's use the following information to very roughly estimate the size of the nucleus. Rutherford, Geiger, and Marsden (his students) fired a beam of alpha particles at a gold foil. The alpha particles had an energy of about 3.70 million electron volts, a charge of +2e, and a mass of about four AMU (one Atomic Mass Unit is 1.66 x 10-27 kg. Gold has atomic number 79.
How close could the alpha particles get to the gold nucleus?
change ev to joules.
work to get close= KE= integral force*dx
the integral is from inf to d, you are looking for d.
KEinJoules= INT kQq/x^2 dx= -kQq/x
and over the limit,
ke in joules= kQq/d solve for d.
Advertising, public
To estimate how close the alpha particles could get to the gold nucleus, we can use Rutherford's scattering formula. This formula relates the impact parameter (b), which is the distance between the path of the alpha particle and the center of the gold nucleus, to the energy (E) and charge (q) of the alpha particle.
The formula is:
b = Z * (q^2) / (4 * π * ε * E)
Where:
b = impact parameter (distance between the path of the alpha particle and the center of the gold nucleus)
Z = atomic number of the gold nucleus (79 in this case)
q = charge of the alpha particle (+2e, where e = elementary charge)
π = Pi (approximately 3.14)
ε = permittivity of free space (approximately 8.85 × 10^-12 F/m)
E = energy of the alpha particle (in joules)
Given:
E = 3.70 million electron volts = 3.70 * 1.6 x 10^-19 J (1eV = 1.6 x 10^-19 J)
q = +2e (where e = 1.6 x 10^-19 C)
Z = atomic number of gold nucleus = 79
Let's calculate the impact parameter (b):
b = 79 * ((2 * 1.6 x 10^-19)^2) / (4 * 3.14 * 8.85 x 10^-12 * 3.70 * 1.6 x 10^-19)
Calculating this equation will give us a rough estimate of how close the alpha particles could get to the gold nucleus.