# Struggling with Uranium-235 Nuclear Fission

• accountdracula
In summary, the conversation was about a question regarding the calculation of the binding energy of a uranium-235 nucleus and the use of different units. The person asking for help initially got an answer of 1.70375 * 10^-3 without units, but after realizing their mistake in using the wrong conversion factor, they were able to correctly calculate the binding energy as 235 MeV or 3.77 x 10^11 joules.
accountdracula

## Homework Statement

I've attached the question I'm having trouble with.

## The Attempt at a Solution

I got an answer of 1.70375 * 10^-3 for the binding energy of a nucleus of uranium 235.
For the third part, "When a uranium-235 nucleus undergoes fission..." I read the value of the binding energy per nucleon for a nucleus half the size of uranium's to be 8.5 *10^-13
So I calculated:
(1.70375 * 10 ^-3) -(2 * 117.5 * 8.5 * 10 ^-13) = difference in total binding energy.

Then I converted this answer in MeV to joules but I got the wrong answer.

#### Attachments

• Uranium 1.jpg
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• Uranium 2.jpg
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Use units! I'm sure that would have made the problem obvious.

accountdracula said:
I got an answer of 1.70375 * 10^-3 for the binding energy of a nucleus of uranium 235.
That quantity is missing units.
What did you get, and what do you expect to get?
accountdracula said:
(1.70375 * 10 ^-3) -(2 * 117.5 * 8.5 * 10 ^-13)
The two different parts seem to use different units. That cannot work.

The values on the vertical axis are already in MeV. Why do you need to convert? Can you show how you get these numbers?

mfb said:
Use units! I'm sure that would have made the problem obvious.

That quantity is missing units.
What did you get, and what do you expect to get?
The two different parts seem to use different units. That cannot work.

I did 7.25 MeV per nucleon * 235 nucleons. I've just realized that I used the wrong conversion factor.
This gives me 1.70375 * 10 ^ 9 eV / 1703. 75 MeV
Nasu, I needed to convert to eV because the question asked for it that way.

Yes, now it makes sense. Your 10^(-3) did not look OK.

Yeah, with that value I calculated:

(8.25 * 235) - 1703.75 = 235 MeV

235 MeV = 3.77 x 10 ^ 11
!
Hurrah!

## 1. What is Uranium-235 nuclear fission?

Uranium-235 nuclear fission is a type of nuclear reaction that involves splitting the nucleus of the uranium-235 isotope into smaller nuclei, resulting in the release of a large amount of energy.

## 2. How is Uranium-235 used for nuclear fission?

Uranium-235 is used as the fuel for nuclear fission reactions because it is a fissile material, meaning it can undergo fission and sustain a chain reaction. It is also relatively abundant and can be enriched to increase its concentration in fuel rods.

## 3. What are the challenges of working with Uranium-235 for nuclear fission?

One of the main challenges with working with Uranium-235 for nuclear fission is its radioactivity. This requires strict safety protocols and specialized equipment to handle and contain the material. Additionally, the waste products of nuclear fission can be highly radioactive and must be properly stored to prevent harm to the environment and living beings.

## 4. How do scientists control the fission process of Uranium-235?

Scientists use various methods to control the fission process of Uranium-235, including controlling the concentration of fuel in the reactor, using control rods to absorb excess neutrons, and monitoring the temperature and pressure within the reactor. These measures help maintain a steady and safe nuclear reaction.

## 5. What are the potential benefits and risks of using Uranium-235 for nuclear fission?

The potential benefits of using Uranium-235 for nuclear fission include its high energy output, which can be used to generate electricity, and its relatively low emissions compared to other energy sources. However, the risks include the potential for accidents, nuclear waste disposal issues, and the potential for the material to be used in weapons if not properly controlled and regulated.

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