Uranium-233 decays to thorium-229 by α decay, but the emissions have different energies and products: 83% emit an α particle with an energy of 4.816 MeV and give ²²⁹Th in its ground state; 15% emit an a particle of 4.773 MeV and give ²²⁹Th in excited state I; and 2% emit a lower energy α particle and give ²²⁹Th in the higher excited state II. Excited state II emits a γ ray of 0.060 MeV to reach excited state I.(b) Find the energy of the a particle that would raise ²³³U to excited state II.

Radioactive new Clyde's give off energy as they decayed to form more stable nucleus is, and this can happen by giving off either particles or energy. So if, for example, we have plutonium isotope in this case P u 2 42 we write it nuclear notation, which has the mass number of high and the atomic number as the sub script. The atomic number is found on the periodic table.

We're told that this decays to produce uranium 2 38 which has a mass number of 92 from the periodic table, plus an Alpha particle which has the symbol of a helium with a mass of four. This decay gives off 4.85 three mega electron volts of energy. The nucleus that forms the uranium nucleus decays further through gamma decay. Given decay does not change the mass or the atomic number of the particle, so it produces The exact same nucleus, but it gives off gamma radiation.

In the symbol for gamma is the Greek symbol gamma. This gives off energy in the form of electromagnetic radiation, specifically a radiation with a wavelength of point 0 to 7 five seven nanometers, we can find the energy that corresponds with that wavelength. Using the equation e equals h time. See over Lambda, where H is a value called Planck's Constant, which is 6.626 times 10 to the minus 34 meter squared times kilograms.

Per second I don't see is the speed of light for three times 10 to the eighth meters per second on our wavelength has to be in meters. So we take our wavelength in nanometers and we know that for every meter there are 10 to the nine nanometers. So this is equivalent to two point 757 times 10 to the minus 11 meters. So we can find the energy of the wavelength in Jules by using those numbers or our energy and Jules.

This 7.21 times 10 to the minus 15. Jules. So the gamma radiation that's given off is equivalent to that value. If we were to see the value of the energy given off going directly from the plutonium directly to the stable nucleus, we would simply add together the two energies because the first step of given in mega electron volts, we convert this to Jules my first changing to electron volts and then using the relationship that every 11 every Jules is equivalent to 6.24 times 10 to the 18 electron bowls.

So the first step produces amount of energy equivalent to 7.78 times 10 to the minus 13 Jules. So if we add these two numbers together, that would give us the overall energy of the process or 7.85 times 10 to the minus 13 Jules would be released in total.

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