In 1910 Rutherford performed a classic experiment in which he directed a beam of alpha particles at a thin gold foil. He unexpectedly observed a few of the particles scattered almost directly backward. This result was not consistent with then current models of atomic structure and led Rutherford to propose the existence of a very dense concentration of positive charge at the center of an atom—the atomic nucleus. The alpha particle has a charge of +2e and the gold nucleus a charge of +79e. Suppose that an alpha particle is initially a great distance from the gold, has a kinetic energy of 3.11 MeV (3.11 106 eV), and is headed directly at a gold nucleus. How close will the particle come to the center of the nucleus? Treat the nucleus and the alpha particle as point charges.

To solve this problem, we can use the principles of electrostatics and conservation of energy. Here's how you can approach it:

Step 1: Convert the given energy of the alpha particle to joules
The energy of the alpha particle is given as 3.11 MeV. Since 1 electron volt (eV) is equal to 1.6 × 10^-19 Joules (J), we can convert the energy by multiplying 3.11 × 10^6 eV by 1.6 × 10^-19 J/eV.

(3.11 × 10^6 eV) × (1.6 × 10^-19 J/eV) = 4.976 × 10^-13 J

Step 2: Use the principle of conservation of energy
According to the principle of conservation of energy, the initial kinetic energy of the alpha particle will be equal to the sum of its final kinetic energy and the electric potential energy.

Kinetic Energy_initial = Kinetic Energy_final + Electric Potential Energy

Step 3: Find the final kinetic energy of the alpha particle
Since the alpha particle is initially very far away from the gold nucleus, we can assume its final kinetic energy is zero because it comes to a halt at the closest point.

Kinetic Energy_final = 0

Step 4: Find the electric potential energy
The electric potential energy between two charges is given by the equation:

Electric Potential Energy = (k * q1 * q2) / r

Where k is the electrostatic constant (8.99 × 10^9 N m^2/C^2), q1 and q2 are the charges, and r is the separation between the charges.

In this case, the charges are +2e for the alpha particle and +79e for the gold nucleus. The separation distance, r, is the distance the alpha particle comes closest to the center of the nucleus.

Step 5: Set up the conservation of energy equation
Using the information from steps 2, 3, and 4, we can set up the equation:

4.976 × 10^-13 J = 0 + (k * (2e) * (79e)) / r

Step 6: Solve for the separation distance, r
Rearrange the equation from Step 5 to solve for r:

r = (k * (2e) * (79e)) / (4.976 × 10^-13 J)

Plug in the values for the electrostatic constant (k = 8.99 × 10^9 N m^2/C^2) and the elementary charge (e = 1.6 × 10^-19 C) to calculate the separation distance, r.

r = (8.99 × 10^9 N m^2/C^2 * (2 * 1.6 × 10^-19 C) * (79 * 1.6 × 10^-19 C)) / (4.976 × 10^-13 J)

Simplifying this expression using the given charges and constants will give you the final answer for how close the alpha particle will come to the center of the gold nucleus.

To find how close the alpha particle will come to the center of the gold nucleus, we can use the conservation of mechanical energy.

1. Convert the kinetic energy of the alpha particle from MeV to joules:
1 MeV = 1.6 x 10^-13 J (conversion factor)
Therefore, 3.11 MeV = 3.11 x 1.6 x 10^-13 J = 4.976 x 10^-13 J.

2. Given that the alpha particle has a charge of +2e and the gold nucleus has a charge of +79e, we can calculate the electrostatic potential energy:
The electrostatic potential energy (U) is given by the equation:
U = (k * |q1 * q2|) / r
where k is Coulomb's constant (8.99 x 10^9 N m^2/C^2), q1 and q2 are the charges of the alpha particle and the gold nucleus respectively, and r is the distance between them.

Plugging in the values, we have:
U = (8.99 x 10^9 N m^2/C^2) * (2e * 79e) / r

3. The initial kinetic energy of the alpha particle is equal to the change in potential energy due to the attractive force between the particles. Therefore, we can set the initial kinetic energy equal to the absolute value of the change in potential energy:
4.976 x 10^-13 J = |Uf - Ui|

4. We can now solve for the final potential energy (Uf) when the alpha particle is at its closest distance to the gold nucleus.

5. Rearranging the equation, we get:
Ui - Uf = 4.976 x 10^-13 J

6. Plugging in the values, we have:
(8.99 x 10^9 N m^2/C^2) * (2e * 79e) / r - Uf = 4.976 x 10^-13 J

7. Since the alpha particle is initially at a great distance from the gold nucleus, its potential energy can be considered zero. Therefore, the equation simplifies to:
(8.99 x 10^9 N m^2/C^2) * (2e * 79e) / r = 4.976 x 10^-13 J

8. Solving for r, we have:
r = (8.99 x 10^9 N m^2/C^2) * (2e * 79e) / (4.976 x 10^-13 J)

9. Plugging in the values and simplifying, we find:
r ≈ 2.7 x 10^-14 m

Therefore, the alpha particle will come as close as approximately 2.7 x 10^-14 meters to the center of the gold nucleus.

Ei = 3.11 MeV

Ei = Ef

Ef = kqQ/r => r = kqQ/Ef
k = 8.99e09 Nm^s/C
e = 1.6e-19 C
1MeV = 1.6e-19 C

r = 7.3076e-14 m