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Problem 41 Hard Difficulty

A device used in radiation therapy for cancer contains 0.50 $\mathrm{g}$
of cobalt $\frac{60}{27} \mathrm{Co}(59.933819 \mathrm{u}) .$ The half-life of $_{27}^{60} \mathrm{Co}$ is 5.27 $\mathrm{yr.}$ Determine the activity of the radioactive material.

Answer

$2.1 \times 10^{13} \mathrm{Bq}$

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Video Transcript

so first weaning to find, um, the number of atoms in our sample because the formula for the decay or for the rate the case is going to land the end. So first thing we need to find Lambda and we'll need to find an so n. It's going to be equal to the mass divided by the mass of each item. So we're going to tank. Our are 0.50 grams, 0.5 zero grounds and there are 1000 um grams, her kilogram. It's only temporary kilograms and then divided by so our atomic mass of 59.9 33 819 You and there are, um, 1.6605 times 10 to the native, 27 kilograms per atomic mass unit. So when we must buy all this out, we get 5.0 times 10 to the 21st Adams. So now we need to find a take a constant or decay constant. Lambda is equal to, um Ln of two divided by the half life So the half life is in 5.27 years. We'll need to convert that two seconds. There are 365 days per year, 24 hours per day and then 3600 seconds per hour. So this Hey, hold divides Ln of two. So this gives us a decay constant of 4.2 times 10 to the negative nine. Inverse science. So finally, we just wanted by the two numbers to get our answer in Beka rails. So and multiplying these two numbers, we now get our final answer two points one times 10 to the 13 Becker rails.