# I knowing what i'm looking for -- alpha particle colliding with lead nucleus

• Eric Peraza
In summary, the conversation is discussing a problem involving an alpha particle with kinetic energy of 15.0MeV colliding with a lead nucleus. The alpha particle is not aimed at the center of the nucleus and has an initial nonzero angular momentum. The distance of closest approach is being asked and it is mentioned that the atomic number of lead is 82 and the alpha particle is a helium nucleus with atomic number 2. The problem involves both mechanics and electromagnetic principles and it is mentioned that the angular momentum is conserved. Possible methods for solving the problem are suggested, including considering the total energy of the system and using regular energy conservation.
Eric Peraza

## Homework Statement

An alpha particle with kinetic energy 15.0MeVmakes a collision with lead nucleus, but it is not "aimed" at the center of the lead nucleus, and has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude L=p0b, where p0 is the magnitude of the initial momentum of the alpha particle and b=1.30×10−12m . (Assume that the lead nucleus remains stationary and that it may be treated as a point charge. The atomic number of lead is 82. The alpha particle is a helium nucleus, with atomic number 2.)

What is the distance of closest approach?

## Homework Equations

What is going on? This is an E&M class and this doesn't seem to have to do with potentials.
any help on where to start or anything would be great.

wow, thorough expectations here.
Is this angular momentum conserved? if so, what L does it have at closest approach?
So, how much KE does it have at closest approach? (no, L2/2mR2 is not zero)
hmm ... where did the REST of the KE go?

what does it mean by closest approach? that's where i mainly got thrown off. and do i calculate the change in voltage and then calculate the change in energy? I am not sure what the question is asking by "closest approach" but my guess is something to do with voltage and energy

Eric Peraza said:
what does it mean by closest approach? that's where i mainly got thrown off. and do i calculate the change in voltage and then calculate the change in energy? I am not sure what the question is asking by "closest approach" but my guess is something to do with voltage and energy

It's a mechanics problem as well as an EM problem. Since you are assuming the lead nucleus is stationary, the angular momentum of the alpha particle is conserved. At closest approach the velocity of the particle is perpendicular to the displacement vector connecting it with the nucleus. Matching that with the given initial angular momentum gives you a relation between v and r. Considering the energy lost by the particle due to the potential gives you another. Solve them simultaneously.

Hey there Eric. If you are still having trouble with this problem, you could consider the total energy of the system. In polar coordinates (where r is the distance of the alpha particle from the lead nucleus), the kinetic energy of the system can be written as, noting that the mass is relativistic (as the alpha particle has 15MeV of energy):

KE = 1/2*m*r(dot)^2 + J^2/(2*m*r^2)

As we can treat this as a central force problem, the potential is given by:

U = -(k*q1*q2)/r

Summing these two together gives:

E = 1/2*m*r(dot)^2 + J^2/(2*m*r^2) - (k*q1*q2)/r

Now, what do we know about r at the distance of closest approach? How can we use this to find r?

Let me know if you are still stuck at this point, but I hope that helps.

PS. Sorry for the lack of latex formatting, I am still learning it!

15 MeV would be relativistic for an electron, but not for a proton or alpha.
you can use regular Energy conservation (quadratic)

## What is an alpha particle colliding with a lead nucleus?

An alpha particle is a type of nuclear radiation that consists of two protons and two neutrons. It is commonly emitted by radioactive materials. A lead nucleus is the central part of an atom of the element lead, which has a high atomic number and is known for its ability to absorb radiation.

## What happens when an alpha particle collides with a lead nucleus?

When an alpha particle collides with a lead nucleus, it can cause several different reactions depending on the energy and direction of the collision. The most common reaction is for the alpha particle to be absorbed by the lead nucleus, which results in the formation of a new, more stable nucleus. This process is known as alpha decay.

## What is the significance of studying alpha particle collisions with lead nuclei?

Studying alpha particle collisions with lead nuclei is important for understanding the behavior and properties of nuclear radiation. It can also provide insight into the structure and composition of atomic nuclei, which is crucial for many areas of science and technology, such as nuclear energy and medicine.

## How are alpha particle collisions with lead nuclei studied?

Alpha particle collisions with lead nuclei are primarily studied using particle accelerators, which can provide the necessary high energies and controlled conditions for these collisions to occur. Scientists also use detectors and other specialized equipment to measure and analyze the particles and radiation produced during these collisions.

## What are the potential risks of alpha particle collisions with lead nuclei?

Alpha particle collisions with lead nuclei can pose a risk to human health if they occur in a high enough concentration or if the particles have enough energy to cause damage to living cells. This is why proper safety protocols and precautions are necessary when conducting experiments involving nuclear radiation. However, the risks can be minimized by following strict guidelines and regulations.

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