# How Do Proton and Alpha-Particle Speeds Affect Their Closest Approach Distance?

• erinec
In summary, the distance of closest approach between a proton and an alpha-particle fired at each other from far away with an initial speed of 0.01c is calculated using the equation for kinetic energy and potential energy of the system. The correct answer is 19.2 fm, but by plugging in the given values for masses, charge, and physical constants, the calculated answer is on the order of 10-16 m. It is suggested to check for any errors in plugging in the values.
erinec

## Homework Statement

A proton and an alpha-particle are fired directly toward each
other from far away each with an initial speed of 0.01c.
Determine their distance of closest approach measured between
their centres.

See below.

## The Attempt at a Solution

What I did was..

Kinetic Energy of the system = Potential Energy of the system
(.5*mp*v2) + (.5*mHe*v2) = k*qp*qHe / r
and then I solved for r, which gave me an incorrect answer.

What am I doing wrong?

I am just wondering.. could you please try this question and tell me what you get?

I get 1.16*105...
which is not the correct answer..

The correct answer is 19.2 fm.;;;

erinec said:

## Homework Statement

A proton and an alpha-particle are fired directly toward each
other from far away each with an initial speed of 0.01c.
Determine their distance of closest approach measured between
their centres.

See below.

## The Attempt at a Solution

What I did was..

Kinetic Energy of the system = Potential Energy of the system
(.5*mp*v2) + (.5*mHe*v2) = k*qp*qHe / r
and then I solved for r, which gave me an incorrect answer.

What am I doing wrong?

I am just wondering.. could you please try this question and tell me what you get?

I get 1.16*105...
which is not the correct answer..

The correct answer is 19.2 fm.;;;
Let's start by giving the relevant masses:
$$m_{He} = 6.64 \times 10^{-27} kg$$
$$m_{p} = 1.67 \times 10^{-27} kg$$
Charge:
$$q_{He} = 3.2 \times 10^{-19}C$$
$$q_p = 1.6 \times 10^{-19} C$$
Physical Constants:
$$k =9\times 10^9 \frac{N m^2}{C^2}$$
$$c = 3\times 10^8 m/s$$
Doing some algebra and plugging in these values I get something on the order of 10-16 m. Are you sure you plugged everything in correct?

I would like to first clarify the question and make sure that the given information is accurate. Is the initial speed of 0.01c referring to 0.01 times the speed of light, or is it 0.01 times the speed of light squared? This will affect the final answer and it is important to have accurate information in scientific calculations.

Assuming that the initial speed is 0.01c (0.01 times the speed of light), I would approach the problem by first calculating the kinetic energy of both particles using the equation KE = (1/2)mv^2. Since the masses of a proton and an alpha-particle are known, we can calculate their kinetic energies.

Next, I would use the conservation of energy principle, which states that the total energy of a system remains constant. This means that the initial kinetic energy of the particles is equal to the final potential energy of the system when they reach their closest approach. Using the equation for electric potential energy, we can solve for the distance of closest approach (r) by setting the initial kinetic energy equal to the potential energy and solving for r.

However, in this case, we also need to consider the relativistic effects due to the high speed of the particles. This can be done by using the relativistic kinetic energy equation KE = (mc^2)/(sqrt(1-(v/c)^2)-1). This will give us a more accurate value for the kinetic energy of the particles.

Once we have the correct values for the kinetic energy, we can proceed with solving for the distance of closest approach. The final answer should be in femtometers (fm), which is a unit used to measure atomic and nuclear distances.

In conclusion, it is important to carefully consider all the given information and use the appropriate equations to solve this problem. Additionally, it is always helpful to double-check the calculations and make sure that the units are consistent throughout the calculation.

## 1. What is electric potential energy?

Electric potential energy is the energy that an electric charge possesses due to its position in an electric field. It is a form of potential energy that is associated with the separation of positive and negative charges.

## 2. How is electric potential energy different from electric potential?

Electric potential energy is a measure of the energy that a charge possesses, while electric potential is a measure of the potential energy per unit charge at a specific point in an electric field.

## 3. What factors affect the amount of electric potential energy?

The amount of electric potential energy is affected by the magnitude of the charges involved, the distance between the charges, and the electric field strength between the charges.

## 4. How is electric potential energy related to work?

Electric potential energy is related to work through the equation: W = qΔV, where W is the work done, q is the charge, and ΔV is the change in electric potential. This means that work is required to move a charge against an electric field, and this work is stored as potential energy.

## 5. How is electric potential energy used in real-world applications?

Electric potential energy is used in a variety of real-world applications, such as in batteries, which store chemical energy and convert it into electric potential energy. It is also used in power plants, where potential energy is converted into kinetic energy to generate electricity. Additionally, electric potential energy is used in capacitors, which store energy in an electric field and are used in electronic devices.

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