A proton is fired from far away toward the nucleus of a mercury atom. Mercury is element number 80, and the diameter of the nucleus is 14.0 fm. The proton is fired at a speed of 1.80×107 m/s. When it passes the nucleus, how close will the proton be to the surface of the nucleus? Assume the nucleus remains at rest.

KE = Fd (work required to stop the proton)

1/2 m v^2 = kq(80q)/r
elementary charge is 1.6e-19 but I don't know the mass of a proton offhand.

mass pf proton is 1.67x10^-27 kg

To determine how close the proton will be to the surface of the nucleus, we need to calculate the distance of closest approach using the conservation of momentum and the Coulomb force.

Step 1: Calculate the momentum of the proton:
Momentum (p) = Mass (m) × Velocity (v)
Given:
Mass of proton (m) = 1.67 × 10^-27 kg
Velocity of proton (v) = 1.80 × 10^7 m/s

p = m × v
p = (1.67 × 10^-27 kg) × (1.80 × 10^7 m/s)
p = 3.006 × 10^-20 kg⋅m/s

Step 2: Calculate the electric force between the proton and the nucleus:
The electric force (Fe) between two charged objects can be calculated using Coulomb's Law:
Fe = (k × q1 × q2) / r^2

Given:
Element number (Z) = 80 (for mercury)
Charge of proton (q1) = +1.6 × 10^-19 C (elementary charge)
Charge of mercury nucleus (q2) = +80 × (1.6 × 10^-19 C) = +1.28 × 10^-17 C
Distance of closest approach (r) = ?

We know that Fe and momentum are conserved, so we can set them equal to each other:
Fe = p^2 / (2m)
(k × q1 × q2) / r^2 = p^2 / (2m)

Simplifying the equation:
(r^2 / k) = (2m × p^2) / (q1 × q2)
r^2 = (2m × p^2) / (k × q1 × q2)

Step 3: Plug in the values and calculate the distance of closest approach:
The Coulomb constant (k) = 8.99 × 10^9 N⋅m^2/C^2

Plugging in the values:
r^2 = (2 × (1.67 × 10^-27 kg) × (3.006 × 10^-20 kg⋅m/s)^2) / ((8.99 × 10^9 N⋅m^2/C^2) × (1.6 × 10^-19 C) × (1.28 × 10^-17 C))

r^2 = 9.2346 × 10^-31 m^2
Taking the square root of both sides:
r = 3.04 × 10^-16 m

Therefore, when the proton passes the nucleus, it will be approximately 3.04 × 10^-16 meters away from the surface of the nucleus.

To find out how close the proton will be to the surface of the nucleus when it passes, we can use the principles of classical mechanics and Coulomb's law.

First, let's calculate the initial kinetic energy of the proton. Kinetic energy (KE) is given by the equation KE = (1/2)mv^2, where m is the mass of the proton and v is its velocity.

The mass of a proton is approximately 1.67 × 10^-27 kg, so the initial kinetic energy is KE = (1/2)(1.67 × 10^-27 kg)(1.8 × 10^7 m/s)^2.

Next, let's calculate the potential energy between the proton and the mercury nucleus. The potential energy (PE) due to electrostatic interaction is given by the equation PE = k * (q1 * q2) / r, where k is Coulomb's constant (8.99 × 10^9 Nm^2/C^2), q1 and q2 are the charges of the two particles, and r is the distance between them.

Since the charge of a proton is 1.6 × 10^-19 C and the charge of a mercury nucleus is +80e, where e is the elementary charge (1.6 × 10^-19 C), the potential energy can be calculated as PE = (8.99 × 10^9 Nm^2/C^2) * (1.6 × 10^-19 C * 80e) / (2 * 14.0 fm).

Now, using the conservation of mechanical energy principle, we can equate the initial kinetic energy to the potential energy when the proton passes by the nucleus. Therefore, KE = PE.

(1/2)(1.67 × 10^-27 kg)(1.8 × 10^7 m/s)^2 = (8.99 × 10^9 Nm^2/C^2) * (1.6 × 10^-19 C * 80e) / (2 * 14.0 fm)

By rearranging the equation and solving for r, we can find the distance from the nucleus surface where the proton will pass:

r = (8.99 × 10^9 Nm^2/C^2) * (1.6 × 10^-19 C * 80e) / [(1/2)(1.67 × 10^-27 kg)(1.8 × 10^7 m/s)^2]

Evaluating this expression will give you the distance from the nucleus surface.

Please note that this calculation assumes classical mechanics and Coulomb's law, so it may not be accurate at very small scales or high energies.