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Problem 56 Easy Difficulty

Many radioisotopes have important industrial, medical, and research applications. One of these is $^{60} \mathrm{Co},$ which has a half - life of 5.2 years and decays by the emission of a beta particle (energy 0.31 MeV) and two gamma photons (energies 1.17 MeV and 1.33 MeV). A scientist wishes to prepare a
$^{60} \mathrm{Co}$ sealed source that will have an activity of at least 10 Ci after 30 months of use. What is the minimum initial mass of $^{60} \mathrm{Co}$ required?

Answer

1.2 \times 10^{-5} \mathrm{kg}

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Top Physics 103 Educators
Marshall S.

University of Washington

Aspen F.

University of Sheffield

Jared E.

University of Winnipeg

Meghan M.

McMaster University

Video Transcript

{'transcript': "for a quick question. We're just We're gonna consider cobalt isotopes 60. So a here is equal to 60 that has 1/2 life of 5.3 years. So t 1/2 is 5.3 years and I convert that into seconds. 1.67 times 7 to 10. The mass of our sample M is 100.4 times 10 to the minus three kilograms. I converted that from Graham's two kilograms, and the mass of a proton, which I call him, said P is 1.66 10% of the minus 27 kilograms for part A were asked to find the decay constant for the isotope. So Lamda here is the decay constant. It's equal to the natural log of two divided by the half life t 1/2 and in units of seconds, this comes out to equal 4.1 floor times 10 to the minus nine seconds. Uh, it wouldn't just be second to would be inverse seconds. Since t 1/2 is on the bottom making box set in as their solution to a part. B wants to wants us to find the number of Adam's that we have so in here is gonna be the number of atoms. And it's equal to the mass of our sample in divided by a times the mass of a proton because a here's the number of nuclei we have in the nucleus. And so we're gonna multiply that by the mess of proton, which is what makes up the nuclei. We find that this is equal to 4.2 times 10 to the 20 Adams. We could go ahead and box it in is their solution for part B. Carsey says that it wants us to find the decays, the number of the case per second, which we call the activity. So the activity we're gonna do know as Delta in over Delta T this is equal to Lambda Times in So what we found for part am part be plugging those values and we find that this is equal to 1.67 times 10 to the 12 decays per second. But decays per second is known as a book where oh, so we're just gonna write this as B Q for becquerels, which is also the again the same as the case per second. So that is our solution to see. Then lastly, Part D were asked to convert this to case per second in two units of curry named after Marie Curry here. So now we have Delta in over Delta T. We're gonna take that value that we found in part C 1.67 times 10 to the 12 becquerels. We're gonna multiply it by the conversion between Curries and becquerels. So one curry abbreviated with C I is equal to three 0.7 times 10 to the 10 becquerels. Okay, So plugging in what we found in C into this expression and carrying out the operation, we find that this is equal to about 45 curry we can box it in Is their solution for part d okay."}

University of Kansas
Top Physics 103 Educators
Marshall S.

University of Washington

Aspen F.

University of Sheffield

Jared E.

University of Winnipeg

Meghan M.

McMaster University