# A Proton is fired from far away towards the nuclues of a mercury Atom

• Blitzp22
In summary, a proton fired at a speed of 32100000 m/s towards the nucleus of a mercury atom, with a diameter of 14.0 fm, will have its closest approach to the surface of the nucleus determined by the equation 1/2 m v^2 = kq1q2/r, where q1 and q2 are the charges on the nucleii. The separation distance, r, can be solved for using this equation.
Blitzp22

## Homework Statement

A proton is fired from far away towards the nucleus of a mercury atom. Mercury is element number 80, and the diameter of the nucleus is 14.0 fm. If the proton is fired at a speed of 32100000 m/s, what is its closest approach to the surface of the nucleus (in fm)? Assume that the nucleus remains at rest.

## Homework Equations

U=kq1q2/r
kinematic equation
K = 1/2 m v^2

## The Attempt at a Solution

So I got this far
1/2(1.67E-27)(3.21E7)^2=(9E9)(q1)(q2)/7fm

but I don't know what to do about the q1 and q2 to get my final answer. Hmmmm toughy

The q's are the charges on the nucleii. You can look them up on a periodic table. Remember, the charge shown on the table is in terms of electron or proton charges which you must convert into the standard unit of Coulombs. No doubt you have the charge of an electron in one of your lists.

Your calc is right, but difficult for me or your teacher to understand. Maybe you, too. You really should start with a general statement that shows your method of attack. Something like this:

Kinetic energy far out is entirely converted to potential energy at the closest approach
or KE far out = PE @ turnaround

yeah but in response to Delphi51's comments, then what are searching for? you would've filled all the unknowns in the equation

Welcome to PF, richnut.
You would still be looking for the separation distance, which is the r in the original post. It just makes more sense to write:
KE far out = PE @ turnaround
1/2 m v^2 = kq1q2/r
The first step clarifies what the second means. Then you go on to solve for r.

I would first clarify the assumptions made in the problem. The problem states that the proton is fired at the nucleus of the mercury atom, but it does not specify the direction of the proton's motion. Is it directly towards the nucleus or at an angle? This could affect the closest approach distance.

Assuming that the proton is fired directly towards the nucleus, we can use the equation for electric potential energy (U) to determine the closest approach distance. We know the mass and speed of the proton, and we can calculate the electric potential energy (U) using the given charge of the proton (q1) and the charge of the nucleus (q2). We can then set this potential energy equal to the kinetic energy of the proton (K) at its closest approach, which can be calculated using the kinematic equation. This will give us the distance (r) at which the proton reaches its closest approach to the nucleus.

To calculate the electric potential energy, we can use the formula U = kq1q2/r, where k is the Coulomb constant, q1 is the charge of the proton (1.6 x 10^-19 C), and q2 is the charge of the nucleus (unknown). Rearranging this equation, we get r = kq1q2/U.

Next, we can calculate the kinetic energy of the proton at its closest approach using the formula K = 1/2mv^2, where m is the mass of the proton (1.67 x 10^-27 kg) and v is its speed (32.1 x 10^6 m/s). We can then set this kinetic energy equal to the electric potential energy (U) and solve for r.

Once we have the value for r, we can convert it from meters to femtometers (fm) by multiplying by 10^15, since 1 fm = 10^-15 m. The final answer will be the closest approach distance in fm.

## 1. What is a proton?

A proton is a subatomic particle that has a positive charge and is found in the nucleus of an atom.

## 2. How does a proton interact with a mercury atom?

When a proton is fired towards the nucleus of a mercury atom, it will be attracted to the negatively charged electrons surrounding the nucleus. This attraction is due to the electromagnetic force between opposite charges.

## 3. What happens when a proton collides with a mercury atom?

When a proton collides with a mercury atom, it can cause the atom to become unstable and undergo radioactive decay. This can result in the release of particles and energy.

## 4. How does the distance between the proton and mercury nucleus affect the interaction?

The closer the proton is to the mercury nucleus, the stronger the attraction between them will be. This is because the electromagnetic force decreases with distance. Therefore, a proton fired from far away will have a weaker interaction with the nucleus compared to one fired from a closer distance.

## 5. Can a proton ever be fired directly into the nucleus of a mercury atom?

No, a proton cannot be fired directly into the nucleus of a mercury atom. This is because the nucleus is incredibly small and densely packed, making it difficult for a proton to penetrate and reach the center.

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