Rutherford, Geiger, and Marsden (his students) fired a beam of alpha particles at a gold foil. The alpha particles had an energy of about 3.6 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?
when the KE is = PE...
PE=k(Q)q/d where Q is the +charge on the gold nucleus, q is the charge on the alpha particle
To determine how close the alpha particles can get to the gold nucleus, we need to consider the concept of the Rutherford scattering experiment and the physics principles involved.
In the Rutherford scattering experiment, alpha particles were fired at a thin gold foil to study the structure of the atom. The observance of most alpha particles passing through the foil with minimal deflection suggested that the atom is mostly empty space, with a small, dense nucleus at the center.
To calculate how close the alpha particles can get to the gold nucleus, we need to consider the interplay between the electrostatic repulsion between the positive charges of the alpha particle and the gold nucleus, and the momentum of the alpha particle.
1. Coulomb's Law:
Coulomb's law states that the force of electrostatic repulsion between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
F = k * (q1 * q2) / r^2
Where:
F = Force of electrostatic repulsion
k = Coulomb's constant (8.99 x 10^9 N m^2/C^2)
q1 = Charge of the alpha particle (+2e)
q2 = Charge of the gold nucleus (+Ze, where Z is the atomic number, which is 79 for gold)
r = Distance between the particles (unknown)
2. Momentum Conservation:
Momentum is conserved in a collision, so we can equate the initial momentum of the alpha particles to its final momentum when it is closest to the gold nucleus.
Initial momentum: P_initial = m * v
where,
m = mass of the alpha particle (about 4 AMU = 6.64 x 10^-27 kg)
v = velocity of the alpha particle (unknown)
Final momentum: P_final = m * v'
where,
v' = velocity of the alpha particle when closest to the nucleus (unknown)
In the experiment, the alpha particles were shot at the gold foil with a known energy of about 3.6 million electron volts. By considering the conservation of energy, we can find the initial velocity of the alpha particles using the relation:
Initial kinetic energy: KE_initial = (1/2) * m * v^2
where,
m = mass of the alpha particle (about 4 AMU = 6.64 x 10^-27 kg)
v = velocity of the alpha particle (unknown)
3. Equating Force and Centripetal Force:
When the alpha particles are closest to the gold nucleus, the electrostatic repulsion between them and the nucleus provides the centripetal force, keeping the alpha particle in a circular orbit. Equating the force of electrostatic repulsion and the centripetal force gives:
F = m * v'^2 / r
4. Combining Equations:
Now we can combine the equations for Coulomb's law, momentum conservation, and equating the forces to solve for the distance (r) when the alpha particle is closest to the gold nucleus.
k * (q1 * q2) / r^2 = m * v'^2 / r
By rearranging the equation, we get:
r = k * (q1 * q2) / (m * v'^2)
Substituting the given values into the equation (with appropriate unit conversions if necessary) will yield the minimum distance (r) that the alpha particles can get close to the gold nucleus.