Solving a Nuclear Reaction Problem: Alpha Particle & Uranium-232

[josephcollins]

Here I have a problem:

A 232/92 Uranium nucleus emits an alpha particle with kinetic energy=5.32MeV. What is the final nucleus and what is the approximate mass(in units) of the final atom.

I can write the equation for the reaction, this will give 4/2 He and 228/90 Th. If the alpha particle has KE=5.32MeV then is this the energy emitted in the reaction? If this is then I can obtain the change in mass. Is this correct? Any help is welcomed, thx. Joe

 

[quasar987]

Here I have a problem:

A 232/92 Uranium nucleus emits an alpha particle with kinetic energy=5.32MeV. What is the final nucleus and what is the approximate mass(in units) of the final atom.

I can write the equation for the reaction, this will give 4/2 He and 228/90 Th. If the alpha particle has KE=5.32MeV then is this the energy emitted in the reaction? If this is then I can obtain the change in mass. Is this correct? Any help is welcomed, thx. Joe

It is almost correct. The total energy "emited" is the kinetic energy of the alpha particle + its mass energy (mc²).

 

[thepatientmental]

Yes, your approach is correct. The equation for the reaction is:

232/92 U --> 4/2 He + 228/90 Th + energy

The alpha particle has a kinetic energy of 5.32 MeV, which is the energy emitted in the reaction. To find the change in mass, we can use Einstein's famous equation E=mc², where E is the energy, m is the mass, and c is the speed of light.

We know the energy (5.32 MeV) and the speed of light (3x10^8 m/s), so we can rearrange the equation to solve for the change in mass:

Δm = E/c²

Substituting in the values, we get:

Δm = (5.32 MeV) / (3x10^8 m/s)^2

Δm = 5.32 x 10^-13 kg

This is the approximate mass of the final atom, which is 228/90 Th. Keep in mind that this is an approximate value, as there may be some slight variations due to nuclear binding energies.

I hope this helps! Let me know if you have any other questions.