In Rutherford's famous scattering experiments that led to the planetary model of the atom, alpha particles (having charges of +2e and masses of 6.64 × 10−27 kg) were fired toward a gold nucleus with charge +79e. An alpha particle, initially very far from the gold nucleus, is fired at 2.6 107 m/s directly toward the nucleus, as in the figure below. How close does the alpha particle get to the gold nucleus before turning around? Assume the gold nucleus remains stationary.

mv²/2=k•q₁q₂/r

r = 2k•q₁q₂/mv²,
where
k =9•10⁹ N•m²/C²
e =1.6•10⁻¹⁹ C.
q₁=2e,
q₂=79e
m=6.64•10⁻²⁷ kg
v=2.6•10⁶ m/s

To calculate how close the alpha particle gets to the gold nucleus before turning around, we can use the principles of energy conservation and the Coulomb's law.

1. First, we need to determine the initial kinetic energy of the alpha particle. We can use the equation:
Kinetic Energy = (1/2) * mass * velocity^2

Given:
mass of alpha particle (m) = 6.64 × 10^(-27) kg
velocity of alpha particle (v) = 2.6 × 10^7 m/s

Plugging in the values:
Kinetic Energy = (1/2) * (6.64 × 10^(-27) kg) * (2.6 × 10^7 m/s)^2

2. Next, we need to calculate the potential energy at the closest distance of approach. Since the alpha particle is initially very far from the gold nucleus, we can assume its potential energy is zero.

3. At the closest distance of approach, all of the initial kinetic energy is converted into potential energy due to the repulsive force between the positively charged alpha particle and the gold nucleus.

Using the principle of energy conservation:
Initial Kinetic Energy = Potential Energy at closest distance of approach

4. We can calculate the potential energy using Coulomb's law:
Potential Energy = (k * charge1 * charge2) / distance,
where k is the electrostatic constant, charge1 is the charge of the alpha particle, charge2 is the charge of the gold nucleus, and distance is the distance of closest approach.

Given:
charge1 = +2e (2 * elementary charge)
charge2 = +79e (79 * elementary charge)

Rearranging the equation:
distance = (k * charge1 * charge2) / Potential Energy

5. Finally, we can calculate the distance of closest approach by substituting the potential energy value from step 3 into the equation obtained in step 4.

Calculating the values:
k (electrostatic constant) = 8.99 * 10^9 N m^2 / C^2
elementary charge (e) = 1.60 * 10^(-19) C

Plugging in all the values:
distance = (8.99 × 10^9 N m^2 / C^2) * (2 * (1.60 × 10^(-19) C) * 79 * (1.60 × 10^(-19) C)) / (1/2 * (6.64 × 10^(-27) kg) * (2.6 × 10^7 m/s)^2)

6. Calculate the value of the formula in step (5) to find the distance of closest approach.
distance = 2.82 × 10^(-14) m

Therefore, the alpha particle gets to a closest distance of approach of approximately 2.82 × 10^(-14) meters before turning around.

To determine how close the alpha particle gets to the gold nucleus before turning around in Rutherford's scattering experiment, we can use the principles of conservation of energy and angular momentum.

Let's start by considering the initial and final conditions of the alpha particle. Initially, the alpha particle is very far from the gold nucleus and has a certain kinetic energy (K) due to its initial velocity. At the closest approach, the alpha particle comes to a stop, resulting in zero kinetic energy. This implies that the change in kinetic energy is equal to the initial kinetic energy.

Using the conservation of energy, we can equate the initial kinetic energy to the potential energy at the closest approach:

(1/2)mv_initial^2 = (1/4πε₀)(2e)(79e)e^2/r

Here, m represents the mass of the alpha particle, v_initial is its initial velocity, ε₀ is the vacuum permittivity, e is the elementary charge, and r is the distance of closest approach.

Next, let's consider the conservation of angular momentum. Since the gold nucleus is assumed to be stationary, the angular momentum of the system initially is equal to the angular momentum at the closest approach. The initial angular momentum of the alpha particle is given by:

mvr_initial = (2e)r

Where r is the distance to the gold nucleus at the closest approach.

Now we have two equations with two unknowns (r and v_initial) that we can solve simultaneously. Rearranging the equations:

v_initial = √((2e²)/(mr_initial)) (Equation 1)
r = [(1/4πε₀)(2e)(79e)e²] / [(1/2)mv_initial²] (Equation 2)

Substituting the value for e (the elementary charge), the mass of the alpha particle, and the given initial velocity (2.6 × 10^7 m/s) into Equation 1, we can calculate v_initial.

Next, substitute the calculated v_initial into Equation 2 to find the value of r, which represents how close the alpha particle gets to the gold nucleus before turning around.

Note: The given values (mass of alpha particle, charge of alpha particle, charge of gold nucleus, and initial velocity) are required to perform the calculations.