# Uranium 235 Fission Question

• dav1d
In summary: I think the E1g=2.99 x 10^-10 J/u x 1g should be E1g=2.99 x 10^-7 J/u x 1g.In summary, when 1 gram of uranium 235 undergoes fission through the reaction U + n -> Te + Zr + n, the mass change is -0.2u and the calculated energy change is 2.99 x 10^-7 J per uranium nucleus. This is found by using the equation E = mc^2, where the mass of the uranium nucleus is 1.66 x 10^-27 kg and the speed of light is 3 x 10^8 m/s. The
dav1d

## Homework Statement

Another mode in which uranium 235 can undergo fission is U + n -> Te + Zr +n. Calculate the energy change when 1g of uranium 235 undergoes fission in this way. The masses are U=235.04u. n=1.0087u. Te=134.92u. Zr=99.92u.

## The Attempt at a Solution

The mass of the left side is 236.0487u
The mass of the right side is 235.8487u

delta M=-0.2u

E=mc^2
=-0.2u x 9x106 J/kg x(1.66x10^-27kg/u)

E1g=-2.988x10^-21J/u x 1g
(Units have failed me here, please tell me where I went wrong...)

E=mc^2
=-0.2u(1.66x10^-27kg/u)*(3x10^8m/s)²
= 2.99 x 10^-10 Joules per Uranium nucleus.
How many Uranium nucleii fission?

For the speed of light, I used 9 x 10^16 J/kg. Can you show the unit analysis for just squaring 3x10^8 m/s ?

1 gram of U-235, the mass of 1 U-235 nucleus is 235.04u given, so 1gram/235.04u=2.5621772x10^21. Then2.99 x 10^-10 x 2.5621772x10^21 = 7.66091009 × 1011 Joules?

Pick any energy formula, say E = mgh. So Joule = kg*m/s²*m.
and Joule/kg = m/s²*m = m²/s²

In my calc, I have
E=mc^2
=-0.2u(1.66x10^-27kg/u)*(3x10^8m/s)²
The u's cancel out and the answer is in kg*m²/s²
which is Joules as seen in E = ½mv² or E = mc².
Your answer looks good now. You just wrote the exponent 16 as 6 in the first calc and there is some confusion about where the decimal place is.

Your calculations are correct, but the units need to be converted to joules (J) per gram (g). The final answer for the energy released would be -2.988x10^-18 J/g. This represents the amount of energy released per gram of uranium 235 undergoing fission in this way. It is important to note that this is a very small amount of energy, but when multiplied by the large amount of uranium 235 used in nuclear reactions, it can be a significant source of energy. Additionally, this calculation does not take into account any additional energy released from the decay of the daughter nuclei (Te and Zr) or the kinetic energy of the neutrons released. Overall, this is a simplified calculation and the actual energy released may vary depending on the specific conditions of the fission reaction.

## 1. What is Uranium 235 fission?

Uranium 235 fission is a nuclear reaction that occurs when an atom of uranium 235 splits into two smaller atoms, releasing a large amount of energy in the form of heat and radiation.

## 2. How is Uranium 235 fission used?

Uranium 235 fission is used in nuclear power plants to generate electricity. It is also used in nuclear weapons to create a powerful explosion.

## 3. What are the advantages of Uranium 235 fission?

The main advantage of Uranium 235 fission is that it produces a large amount of energy, making it a reliable source of electricity. It also does not produce greenhouse gas emissions, unlike fossil fuels.

## 4. What are the risks associated with Uranium 235 fission?

The main risk of Uranium 235 fission is the potential for accidents and the release of radioactive material, which can have harmful effects on human health and the environment. There is also the risk of nuclear proliferation, where the technology and materials used for Uranium 235 fission can be used to create nuclear weapons.

## 5. Can Uranium 235 fission be replaced with alternative energy sources?

While there are alternative energy sources such as renewable energy, they currently cannot match the energy output of Uranium 235 fission. However, research and development in alternative energy sources continue to progress in order to reduce our reliance on nuclear energy.

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