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Ch.20 - Nuclear Chemistry

Chapter 20, Problem 62

The electronic systems on the New Horizons spacecraft, which launched on January 19, 2006, and reached its closest approach to Pluto on July 14, 2015, were powered by elec-tricity generated by heat. The heat came from the radioac-tive decay of 238Pu in the 11 kg of 238PuO2 fuel onboard. The generator provided 240 W when the spacecraft was launched. If the power output is directly proportional to the amount of 238Pu in the generator, what was the power output when the spacecraft reached Pluto? The half-life of 238Pu is 87.7 y.

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Hello everyone today. We have the following problem. The Chernobyl disaster took place on april 26th, 1986. When reactor number four in the Chernobyl nuclear power plant exploded, the reactor core was exposed and burned in the air for eight days, releasing enormous amounts of radio activity. The reactor was run on uranium 2 35 and had 1.951 times 10 to the 50 kg of fuel when it exploded and the radiation levels in the vicinity for about 20,000 ro engine per hour. If the radiation levels are directly proportional to the amount of uranium 35 what will be the radiation level in the vicinity of the reactor On april 23rd, 2023. The half life of uranium to 35 is 7.38 times 10 to the eighth year. So before we begin in all of this, we have to determine or we have to recall the first order half life formula for radioactive substance. That's going to be the natural log of. This is going to be a ratio of the number of radioactive nuclei at time of T Divided by the initial number of radioactive nuclei in sub zero. And this is going to be equal to negative rates constant times the time. Now that we have that out of the way, we have to also recall our half life formula which is T sub half is equal to 0.693 divided by K. Or our rate constant. And so if we were to substitute the values of K. In the integrated rate law, We would have to first solve for our K. Our K could equal 0. divided by our half life. If we were to substitute this value of K. Into this equation here we would have our natural log and some tea and sub zero is equal to negative 0. times T over our half life. And so it says that the incident took place on April 26 of 1986. And if we were to trace that to 2023, that would be the same date exactly 37 years Later. So 37 years later. So let's place the values into our formula. So we have our natural log about ratio we have our negative 0.693. That was 37 years. Our half life is 7.38 times 10 to the eighth years. Simplifying this, we're going to get our natural log of. That ratio is equal to 3. times 10 to the negative eighth. To get rid of the natural log we just have to raise each of these values to the power of E. Or two E. And if you simplify that, we're gonna get this ratio Is equal to 0.999 repeating. So the radioactive levels in the vicinity of a reactor. We have to solve For the present levels. So we have 20,000 row engine per hour Something Gotta multiply by this .999 repeating. And that's gonna give us 19,999 row engine per hour. So essentially the radiation levels are not going to change much and thus we can say the radiation levels will be about the same. And so that's gonna be our final answer. The radiation levels will be about the same 37 years down the line. And with that we have answered the question overall. I hope this helped. And until next time.
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